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hdoj 4193 Non-negative Partial Sums - 单调队列

y @ The Angry Teletubbies — Tue, 03 Apr 2012 03:56:00 GMT

【题意】：(x��)�l�定一个长度�ؓ(f��)n的环序列�Q�可以�Ş成n�U�链序列�Q�问有多��种铑ֺ�列满��x(ch��ng)��有前i��和都大于等�?(1<=i<=n)�?br />
【题解】：(x��)首先把环序列切断�Q�变成两倍长度。预处理所有前 i ��和�?br /> 枚�D��L(f��ng)��j�Q��用单调队列求区间最��|��如果最��值大于sum[j]则答案加一�?br />
【代码】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 #include "cmath"
8 #include "string"
9 #include "cctype"
10 #include "map"
11 #include "iomanip"
12 using namespace std;
13 #define pb push_back
14 #define mp make_pair
15 #define fi first
16 #define se second
17 #define lc(x) (x << 1)
18 #define rc(x) (x << 1 | 1)
19 #define lowbit(x) (x & (-x))
20 #define ll long long
21 #define maxn 1000050
22 int n;
23 int val[maxn];
24 int sum[maxn<<1];
25 int que[maxn<<1], head, tail;
26
27 int main() {
28     while(~scanf("%d", &n)) {
29         if(!n) break;
30         sum[0] = 0;
31         for(int i = 1; i <= n; i++) {
32             scanf("%d", &val[i]);
33             sum[i] = sum[i-1] + val[i];
34         }
35         for(int i = n + 1; i <= 2 * n; i++) {
36             sum[i] = sum[i-1] + val[i-n];
37         }
38         int ans = 0;
39         head = tail = 0;
40         que[tail++] = 0;
41         for(int i = 1; i <= 2 * n; i++) {
42             if(i > n) {
43                 while(head < tail && que[head] < i - n - 1) head++;
44                 if(sum[que[head]] >= sum[i-n-1]) ans++;
45             }
46             while(head < tail && sum[que[tail-1]] > sum[i]) tail--;
47             que[tail++] = i;
48         }
49         cout << ans << endl;
50     }
51
52     return 0;
53 }
54

y @ The Angry Teletubbies 2012-04-03 11:56 发表评论

hdoj 3415 Max Sum of Max-K-sub-sequence - 单调队列

y @ The Angry Teletubbies — Mon, 02 Apr 2012 13:14:00 GMT

【题意】：(x��)�l�出一个长度�ؓ(f��)n的环状序列，求一个长度不��过k的和最大的�q�箋(hu��)子序列�?br />
【题解】：(x��)很容易想到O(n*k)的做法，但是赤裸裸的TLE�?br /> 先求出所有前n��和�?br /> 枚�D每一个位�|?i 作�ؓ(f��)�l�束点，现在我们要找到最��的sum[j], i - k <= j < i .
朴素做法是直接枚举O(k)�Q�改�q�一点的做法是用�U�段�?w��i)查询区间最大值O(logk).
但是可以证明有单调性，故可以用单调队列优化�Q�查询直接优化到O(1)�?br /> 所以最�l�复杂度是O(n)�?br />
【代码】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 #include "cmath"
8 #include "string"
9 #include "cctype"
10 #include "map"
11 #include "iomanip"
12 using namespace std;
13 #define pb push_back
14 #define mp make_pair
15 #define fi first
16 #define se second
17 #define lc(x) (x << 1)
18 #define rc(x) (x << 1 | 1)
19 #define lowbit(x) (x & (-x))
20 #define ll long long
21 #define maxn 100050
22 const int inf = 1 << 30;
23 int que[maxn<<1], head, tail;
24 int sum[maxn<<1];
25 int val[maxn];
26 int n, k;
27 int ans, s, t;
28 int main() {
29     int T;
30     scanf("%d", &T);
31     while(T--) {
32         scanf("%d%d", &n, &k);
33         sum[0] = 0;
34         for(int i = 1; i <= n; i++) {
35             scanf("%d", &val[i]);
36             sum[i] = val[i] + sum[i-1];
37         }
38         for(int i = n + 1; i <= 2 * n; i++) {
39             sum[i] = sum[i-1] + val[i-n];
40         }
41         ans = -inf;
42         head = tail = 0;
43         que[tail++] = 0;
44         for(int i = 1; i <= 2 * n; i++) {
45             while(head < tail && que[head] < i - k) head++;
46             if(ans < sum[i] - sum[que[head]]) {
47                 ans = sum[i] - sum[que[head]];
48                 s = que[head] + 1, t = i;
49             }
50             while(head < tail && sum[que[tail-1]] > sum[i]) tail--;
51             que[tail++] = i;
52         }
53         if(s > n) s -= n;
54         if(t > n) t -= n;
55         cout << ans << " " << s << " " << t << endl;
56     }
57     return 0;
58 }
59

y @ The Angry Teletubbies 2012-04-02 21:14 发表评论

y @ The Angry Teletubbies — Thu, 08 Mar 2012 10:32:00 GMT

【题意】：(x��)有n个黑帮团伙成员，且有两个帮派 �?br /> D x y 表示 x y属于不同的帮?hu��)z�；
A x y 询问 x y的关�p�R�?br /> 关系有三�U�：(x��)1.Not sure yet. 2.In different gangs. 3.In the same gangs.
�l�出m个操作，回答所有询问�?br />
【题解】：(x��)�q�查集题目，可是想不出来�Q�给队友�Ҏ(gu��)��?ji��n)�?br /> 对于每个人，拆点x x’�?br /> 对于D x y�q�个操作,把x和y’所在的分量合�ƈ�Q�x’和y所在的分量合�ƈ�?br /> 对于A x y�q�个询问,
如果find(x) == find(y)�Q�则他们属于同一个帮?hu��)z�；
否则�Q�如果find(x) == find(y')�Q�他们属于不同帮?hu��)z�；
否则�Q�他们的关系不确定�?br />
最好在�U怸��M��图，�q�样比较�Ҏ(gu��)��理解�?br />
【代码】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 #include "cmath"
8 #include "string"
9 #include "cctype"
10 #include "map"
11 #include "iomanip"
12 using namespace std;
13 #define pb push_back
14 #define lc(x) (x << 1)
15 #define rc(x) (x << 1 | 1)
16 #define lowbit(x) (x & (-x))
17 #define ll long long
18 #define maxn 200050
19 int n, m;
20 int fa[maxn];
21
22 void init() {
23     for(int i = 0; i < maxn; i++) fa[i] = i;
24 }
25
26 int find(int x) {
27     if(x != fa[x]) return fa[x] = find(fa[x]);
28     return fa[x];
29 }
30
31 bool un(int a, int b) {
32     a = find(a), b = find(b);
33     if(a == b) return false;
34     fa[a] = b;
35     return true;
36 }
37
38 int main() {
39     int T, a, b;
40     char op[5];
41     scanf("%d", &T);
42     while(T--) {
43         init();
44         scanf("%d%d", &n, &m);
45         for(int i = 0; i < m; i++) {
46             scanf("%s%d%d", op, &a, &b);
47             if(op[0] == 'A') {
48                 if(find(a) == find(b)) printf("In the same gang.\n");
49                 else if(find(a) == find(b+n)) printf("In different gangs.\n");
50                 else printf("Not sure yet.\n");
51             } else {
52                 un(a, b + n), un(a + n, b);
53             }
54         }
55     }
56     return 0;
57 }
58

y @ The Angry Teletubbies 2012-03-08 18:32 发表评论

poj 1330 Nearest Common Ancestors - LCA + Tarjan��法

y @ The Angry Teletubbies — Fri, 27 Jan 2012 08:35:00 GMT

【题意】：(x��)求一��|��(w��i)上的最�q�公��q��先�?br />
【题解】：(x��)模板题�?br />
【代码】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 #include "cmath"
8 #include "string"
9 #include "cctype"
10 using namespace std;
11 #define pb push_back
12 #define lc(x) (x << 1)
13 #define rc(x) (x << 1 | 1)
14 #define lowbit(x) (x & (-x))
15 #define ll long long
16 #define maxn 10005
17 vector<int> tree[maxn], Q[maxn];
18 int in[maxn], n;
19 bool visit[maxn];
20 int fa[maxn];
21
22 void init() {
23     for(int i = 0; i <= n; i++) fa[i] = i, tree[i].clear(), Q[i].clear();
24     memset(visit, false, sizeof(visit));
25     memset(in, 0, sizeof(in));
26 }
27
28 int find(int x) {
29     if(x == fa[x]) return x;
30     else return fa[x] = find(fa[x]);
31 }
32
33 bool un(int a, int b) {
34     a = find(a), b = find(b);
35     if(a == b) return false;
36     fa[b] = a;
37     return true;
38 }
39
40 void Tarjan(int u) {
41     int size = tree[u].size();
42     for(int i = 0; i < size; i++) {
43         Tarjan(tree[u][i]);
44         un(u, tree[u][i]);
45     }
46     visit[u] = true, size = Q[u].size();
47     for(int i = 0; i < size; i++) {
48         if(visit[Q[u][i]]) {
49             printf("%d\n", find(Q[u][i]));
50             return;
51         }
52     }
53 }
54
55 int main() {
56     int T, u, v;
57     scanf("%d", &T);
58     while(T--) {
59         scanf("%d", &n);
60         init();
61         for(int i = 0; i < n - 1; i++) {
62             scanf("%d%d", &u, &v);
63             tree[u].pb(v);
64             in[v]++;
65         }
66         scanf("%d%d", &u, &v);
67         Q[u].pb(v), Q[v].pb(u);
68         int root = -1;
69         for(int i = 1; i <= n; i++) {
70             if(in[i] == 0) {
71                 root = i;
72                 break;
73             }
74         }
75         Tarjan(root);
76     }
77     return 0;
78 }
79

y @ The Angry Teletubbies 2012-01-27 16:35 发表评论

y @ The Angry Teletubbies — Sat, 21 Jan 2012 14:28:00 GMT

【题意】：(x��)有一块面�U��ؓ(f��)h*w的通知栏，往上面贴一�?*wi的通知上去�Q�脓(chu��ng)的位�|��L��选择可以贴的最高�(sh��)��|�的最左端�Q�且通知不能重叠。给出n个通知�Q��ƈ询问它们分别贴在�W�几行�?br />
【题解】：(x��)�U�段�?w��i)，�l�护�U�段�?w��i)的区间为[1, min(h,n)]�Q�因为最坏情况也��׃��行脓(chu��ng)一个，所以后面的是多余的�?br /> 因�ؓ(f��)查询和更新是同时的，所以我写在一起了(ji��n)�?br />
【代码】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 #include "cmath"
8 #include "string"
9 #include "cctype"
10 using namespace std;
11 #define pb push_back
12 #define lc(x) (x << 1)
13 #define rc(x) (x << 1 | 1)
14 #define lowbit(x) (x & (-x))
15 #define ll long long
16 #define MAX 200050
17 int sum[MAX<<2];
18 int h, w, n;
19
20 void pushup(int p) {
21     sum[p] = max(sum[lc(p)], sum[rc(p)]);
22 }
23
24 void build(int l, int r, int p) {
25     sum[p] = w;
26     if(l == r) return;
27     int mid = (l + r) >> 1;
28     build(l, mid, lc(p));
29     build(mid + 1, r, rc(p));
30 }
31
32 int query(int L, int l, int r, int p) {
33     if(l == r) {
34         sum[p] -= L; // 扑ֈ�最优�(sh��)��|��ƈ更新
35         return l;
36     }
37     int mid = (l + r) >> 1, ans = -1;
38     if(L <= sum[lc(p)]) ans = query(L, l, mid, lc(p));
39     else ans = query(L, mid + 1, r, rc(p));
40     pushup(p);
41     return ans;
42 }
43
44 int main() {
45     int L;
46     while(~scanf("%d%d%d", &h, &w, &n)) {
47         if(h > n) h = n;
48         build(1, h, 1);
49         for(int i = 0; i < n; i++) {
50             scanf("%d", &L);
51             if(sum[1] < L) printf("-1\n");
52             else printf("%d\n", query(L, 1, h, 1));
53         }
54     }
55     return 0;
56 }
57

y @ The Angry Teletubbies 2012-01-21 22:28 发表评论

y @ The Angry Teletubbies — Sun, 15 Jan 2012 06:44:00 GMT

【题意】：(x��)�l�出n个数�l�成的一个序列，我们可以把前m个数�U�d��序列的后面，�q�样我们一共可以得到n个不同的序列。求�q�n个序列中�Q�逆序�Ҏ(gu��)��ؓ(f��)多少�?br />
【题解】：(x��)可以利用�?w��i)状数组或者线�D�|��(w��i)在O(nlogn)旉��里求出初始序列的逆序对，然后每次用O(1)的时间推出每�U�d��一个数到序列后面之后的逆序寏V�?br /> ��d��杂度O(nlogn + n)�?br />
【代码】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 #include "cmath"
8 #include "string"
9 using namespace std;
10 #define pb push_back
11 #define lc(x) (x << 1)
12 #define rc(x) (x << 1 | 1)
13 #define lowbit(x) (x & (-x))
14 #define ll long long
15 #define maxn 5005
16 const int inf = 1 << 30;
17 int sum[maxn], n, val[maxn];
18
19 void update(int x) {
20     for(int i = x; i < maxn; i += lowbit(i))
21         sum[i]++;
22 }
23
24 int SUM(int x) {
25     int res = 0;
26     for(int i = x; i; i -= lowbit(i))
27         res += sum[i];
28     return res;
29 }
30
31 int main() {
32     while(~scanf("%d", &n)) {
33         memset(sum, 0, sizeof(sum));
34         int ans = inf, tmp = 0;
35         for(int i = 0; i < n; i++) {
36             scanf("%d", &val[i]);
37             tmp += SUM(n) - SUM(val[i]);
38             update(val[i]+1);
39         }
40         for(int i = 0; i < n - 1; i++) {
41             tmp += n - 2 * val[i] - 1;
42             ans = min(ans, tmp);
43         }
44         cout << ans << endl;
45     }
46     return 0;
47 }
48

y @ The Angry Teletubbies 2012-01-15 14:44 发表评论

y @ The Angry Teletubbies — Wed, 11 Jan 2012 07:39:00 GMT

【题意】：(x��)�l�出n(n<=100000)个数的初始��|��~�号1~n�Q�再�l�出q(q<=100000)个操作，不定时往�W�i个数到第j个数上加上一个常数c或者查询第i个数到第j个数的和�?br />
【题解】：(x��)�q�是�U�段�?w��i)的入门题目�Q�凭自己的理解裸打了(ji��n)一�ơ线�D�|��(w��i)�Q�对延迟标记理解得更透切�?ji��n)�?br /> 注意要用long long.

【代码】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 using namespace std;
8 #define pb push_back
9 #define lc(x) (x << 1)
10 #define rc(x) (x << 1 | 1)
11 #define lowbit(x) (x & (-x))
12 #define MAX 100050
13 #define ll long long
14 ll lazy[MAX<<2], sum[MAX<<2];
15 int n, q;
16
17 void pushup(int p) {
18     sum[p] = sum[lc(p)] + sum[rc(p)];
19 }
20
21 void pushdown(int p, int len) {
22     if(lazy[p]) {
23         lazy[lc(p)] += lazy[p];
24         lazy[rc(p)] += lazy[p];
25         sum[lc(p)] += (len - (len >> 1)) * lazy[p];
26         sum[rc(p)] += (len >> 1) * lazy[p];
27         lazy[p] = 0;
28     }
29 }
30
31 void build(int l, int r, int p) {
32     lazy[p] = 0;
33     if(l == r) {
34         scanf("%lld", &sum[p]);
35         return;
36     }
37     int mid = (l + r) >> 1;
38     build(l, mid, lc(p));
39     build(mid + 1, r, rc(p));
40     pushup(p);
41 }
42
43 void update(int L, int R, ll c, int l, int r, int p) {
44     if(L <= l && r <= R) {
45         lazy[p] += c;
46         sum[p] += c * (r - l + 1);
47         return;
48     }
49     pushdown(p, r - l + 1);
50     int mid = (l + r) >> 1;
51     if(L <= mid) update(L, R, c, l, mid, lc(p));
52     if(R > mid) update(L, R, c, mid + 1, r, rc(p));
53     pushup(p);
54 }
55
56 ll query(int L, int R, int l, int r, int p) {
57     ll res = 0;
58     if(L <= l && r <= R) {
59         return sum[p];
60     }
61     pushdown(p, r - l + 1);
62     int mid = (l + r) >> 1;
63     if(L <= mid) res += query(L, R, l, mid, lc(p));
64     if(R > mid) res += query(L, R, mid + 1, r, rc(p));
65     pushup(p);
66     return res;
67 }
68
69 int main() {
70     char str[5];
71     int a, b;
72     ll c;
73     while(~scanf("%d%d", &n, &q)) {
74         build(1, n, 1);
75         while(q--) {
76             scanf("%s%d%d", str, &a, &b);
77             if(str[0] == 'Q') printf("%lld\n", query(a, b, 1, n, 1));
78             else {
79                 scanf("%lld", &c);
80                 update(a, b, c, 1, n, 1);
81             }
82         }
83     }
84     return 0;
85 }

y @ The Angry Teletubbies 2012-01-11 15:39 发表评论

y @ The Angry Teletubbies — Wed, 11 Jan 2012 07:12:00 GMT

【题意】：(x��)��L��周知�Q�Dota里面有个屠夫�Q�他有个吊钩。现在考虑他的吊钩由n(n<=100000)�D�小��组成，初始�Ӟ��都是铜棒。给出q(q<=100000)个操作，每个操作都可以把�W�i�D�到�W�j�D늚��棒替换成其他类型的��（铜棒�Q�银��，金棒�Q��h(hu��n)值分别�ؓ(f��)1,2,3�Q�。问�Q�经�q�q个操作之后，�q�个吊钩的��M�h(hu��n)��gؓ(f��)多少�?br />
【题解】：(x��)�U�段�?w��i)，区间替换�Q�区间查询�?br /> 初学�U�段�?w��i)，理解延迟标记是关键�?br /> 本题比较�Ҏ(gu��)��Q�查询只有一�ơ，而且是查询整个区��_(d��)��输出stick[1]卛_��?br />
【代码】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 using namespace std;
8 #define pb push_back
9 #define lc(x) (x << 1)
10 #define rc(x) (x << 1 | 1)
11 #define MAX 100005
12 int stick[MAX<<2], n, m;
13 int lazy[MAX<<2];
14
15 void pushup(int p) {
16     stick[p] = stick[lc(p)] + stick[rc(p)];
17 }
18
19 void pushdown(int p, int len) {
20     if(lazy[p]) {
21         lazy[lc(p)] = lazy[rc(p)] = lazy[p];
22         stick[lc(p)] = (len - (len >> 1)) * lazy[p];
23         stick[rc(p)] = (len >> 1) * lazy[p];
24         lazy[p] = 0;
25     }
26 }
27
28 void build(int l, int r, int p) {
29     lazy[p] = 0;
30     if(l == r) {
31         stick[p] = 1;
32         return;
33     }
34     int mid = (l + r) >> 1;
35     build(l, mid, lc(p));
36     build(mid + 1, r, rc(p));
37     pushup(p);
38 }
39
40 void update(int L, int R, int c, int l, int r, int p) {
41     if(L <= l && r <= R) {
42         lazy[p] = c;
43         stick[p] = c * (r - l + 1);
44         return;
45     }
46     pushdown(p, r - l + 1);
47     int mid = (l + r) >> 1;
48     if(L <= mid) update(L, R, c, l, mid, lc(p));
49     if(R > mid) update(L, R, c, mid + 1, r, rc(p));
50     pushup(p);
51 }
52
53 int main() {
54     int T, Case = 1;
55     int x, y, z;
56     scanf("%d", &T);
57     while(T--) {
58         scanf("%d%d", &n, &m);
59         build(1, n, 1);
60         while(m--) {
61             scanf("%d%d%d", &x, &y, &z);
62             update(x, y, z, 1, n, 1);
63         }
64         printf("Case %d: The total value of the hook is %d.\n", Case++, stick[1]);
65     }
66     return 0;
67 }
68

y @ The Angry Teletubbies 2012-01-11 15:12 发表评论

y @ The Angry Teletubbies — Tue, 10 Jan 2012 07:00:00 GMT

【题意】：(x��)�l�出n个学生的初始成�W�Q�编�?~n�Q�不定时的修�Ҏ(gu��)��个学生的成�W和查询某一�D�连�l�编��h��l�的最高倹{�?n<=200000)

【题解】：(x��)数据�l�构题目�Q�线�D�|��(w��i)�U�了(ji��n)�Q�不知道�?w��i)状数组行不行�?br />
【代码】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 using namespace std;
8 #define pb push_back
9 #define MAX 200050
10 #define lc(x) (x << 1)
11 #define rc(x) (x << 1 | 1)
12 const int inf = 1 << 30;
13 int grade[MAX<<2], n, m;
14
15 void pushup(int p) {
16     grade[p] = max(grade[lc(p)], grade[rc(p)]);
17 }
18
19 void build(int l, int r, int p) {
20     if(l == r) {
21         scanf("%d", &grade[p]);
22         return;
23     }
24     int mid = (l + r) >> 1;
25     build(l, mid, lc(p));
26     build(mid + 1, r, rc(p));
27     pushup(p);
28 }
29
30 int query(int L, int R, int l, int r, int p) {
31     int res = -inf;
32     if(L == l && R == r) return grade[p];
33     int mid = (l + r) >> 1;
34     if(L > mid) res = max(res, query(L, R, mid + 1, r, rc(p)));
35     else if(R <= mid) res = max(res, query(L, R, l, mid, lc(p)));
36     else res = max(res, max(query(L, mid, l, mid, lc(p)), query(mid + 1, R, mid + 1, r, rc(p))));
37     return res;
38 }
39
40 void update(int x, int val, int l, int r, int p) {
41     if(l == r) {
42         grade[p] = val;
43         return;
44     }
45     int mid = (l + r) >> 1;
46     if(x <= mid) update(x, val, l, mid, lc(p));
47     else update(x, val, mid + 1, r, rc(p));
48     pushup(p);
49 }
50
51 int main() {
52     int a, b;
53     char str[10];
54     while(~scanf("%d%d", &n, &m)) {
55         build(1, n, 1);
56         while(m--) {
57             scanf("%s%d%d", str, &a, &b);
58             if(str[0] == 'Q') printf("%d\n", query(a, b, 1, n, 1));
59             else update(a, b, 1, n, 1);
60         }
61     }
62     return 0;
63 }

y @ The Angry Teletubbies 2012-01-10 15:00 发表评论

y @ The Angry Teletubbies — Mon, 09 Jan 2012 17:25:00 GMT

【题意】：(x��)�l�出n个兵营，每个兵营初始都有一定数量的士兵�Q�兵营的士兵数量�?x��)改变，不定时查询第i个兵营到�W�j个兵营的士兵数量�Q�输出每�ơ询问的�l�果�?n <= 50000)

【题解】：(x��)很简单的一道数据结构题目�?br /> �q�种题目如果能用�?w��i)状数组最好就用，因�ؓ(f��)�?w��i)状数组太容易写了(ji��n)�?br /> 最�q�在学线�D�|��(w��i)�Q�所以两�U�版本的都写�?ji��n)一�ơ�?br />
【代码】：(x��)

【树(w��i)状数�l�】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 using namespace std;
8 #define pb push_back
9 #define maxn 50005
10 #define lowbit(x) (x & (-x))
11 int val[maxn], n;
12
13 void add(int x, int e) {
14     for(int i = x; i < maxn; i += lowbit(i)) {
15         val[i] += e;
16     }
17 }
18
19 int sum(int x) {
20     int res = 0;
21     for(int i = x; i; i -= lowbit(i)) {
22         res += val[i];
23     }
24     return res;
25 }
26
27 int main() {
28     int T, Case = 1, e, a, b;
29     char str[10];
30     scanf("%d", &T);
31     while(T--) {
32         scanf("%d", &n);
33         memset(val, 0, sizeof(val));
34         for(int i = 1; i <= n; i++) {
35             scanf("%d", &e);
36             add(i, e);
37         }
38         printf("Case %d:\n", Case++);
39         while(1) {
40             scanf("%s", str);
41             if(str[0] == 'E') break;
42             else {
43                 scanf("%d%d", &a, &b);
44                 if(str[0] == 'Q') printf("%d\n", sum(b) - sum(a - 1));
45                 else if(str[0] == 'A') add(a, b);
46                 else add(a, -b);
47             }
48         }
49     }
50     return 0;
51 }

【线�D�|��(w��i)】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "algorithm"
5 #include "vector"
6 #include "queue"
7 using namespace std;
8 #define pb push_back
9 #define lc(x) (x << 1)
10 #define rc(x) (x << 1 | 1)
11 #define MAX 50005
12 int seg[MAX<<2];
13 int n;
14 void pushup(int p) {
15     seg[p] = seg[lc(p)] + seg[rc(p)];
16 }
17
18 void build(int l, int r, int p) {
19     if(l == r) {
20         scanf("%d", &seg[p]);
21         return;
22     }
23     int mid = (l + r) >> 1;
24     build(l, mid, lc(p));
25     build(mid + 1, r, rc(p));
26     pushup(p);
27 }
28
29 int query(int L, int R, int l, int r, int p) {
30     if(L == l && R == r) return seg[p];
31     int mid = (l + r) >> 1;
32     if(L > mid) return query(L, R, mid + 1, r, rc(p));
33     else if(R <= mid) return query(L, R, l, mid, lc(p));
34     else return query(L, mid, l, mid, lc(p)) + query(mid + 1, R, mid + 1, r, rc(p));
35 }
36
37 void update(int x, int val, int l, int r, int p) {
38     if(l == r) {
39         seg[p] += val;
40         return;
41     }
42     int mid = (l + r) >> 1;
43     if(x <= mid) update(x, val, l, mid, lc(p));
44     else update(x, val, mid + 1, r, rc(p));
45     pushup(p);
46 }
47
48 int main() {
49     int T, Case = 1, a, b;
50     char str[10];
51     scanf("%d", &T);
52     while(T--) {
53         scanf("%d", &n);
54         build(1, n, 1);
55         printf("Case %d:\n", Case++);
56         while(1) {
57             scanf("%s", str);
58             if(str[0] == 'E') break;
59             else {
60                 scanf("%d%d", &a, &b);
61                 if(str[0] == 'Q') printf("%d\n", query(a, b, 1, n, 1));
62                 else if(str[0] == 'A') update(a, b, 1, n, 1);
63                 else update(a, -b, 1, n, 1);
64             }
65         }
66     }
67     return 0;
68 }

y @ The Angry Teletubbies 2012-01-10 01:25 发表评论

poj 2823 Sliding Window - 单调队列

y @ The Angry Teletubbies — Sat, 05 Nov 2011 01:54:00 GMT

【题意】：(x��)�l�出n个数�Q�求每连�l�k个数之间的最大最��倹{�?br />
【题解】：(x��)单调队列入门题�?br />
【代码】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 using namespace std;
5 #define maxn 1000500
6 int n, k, val;
7 int maxx[maxn], minn[maxn];
8 int head1, tail1, head2, tail2, tot;
9 struct Drab {
10     int val, idx;
11     Drab(){}
12     Drab(int _val, int _idx) {
13         val = _val, idx = _idx;
14     }
15 }que1[maxn], que2[maxn];
16
17 void init() {
18     tot = 0;
19     head1 = tail1 = head1 = tail1 = 0;
20 }
21
22 void push1(int val, int idx) {
23     while(head1 < tail1 && que1[tail1 - 1].val > val) tail1--;
24     que1[tail1++] = Drab(val, idx);
25 }
26
27 void push2(int val, int idx) {
28     while(head2 < tail2 && que2[tail2 - 1].val < val) tail2--;
29     que2[tail2++] = Drab(val, idx);
30 }
31
32 int main() {
33     while(~scanf("%d%d", &n, &k)) {
34         init();
35         for(int i = 0; i < n; i++) {
36             scanf("%d", &val);
37             push1(val, i), push2(val, i);
38             if(i >= k - 1) {
39                 if(que1[head1].idx <= i - k) head1++;
40                 minn[tot] = que1[head1].val;
41                 if(que2[head2].idx <= i - k) head2++;
42                 maxx[tot] = que2[head2].val;
43                 tot++;
44             }
45         }
46         for(int i = 0; i < tot; i++) printf("%d ", minn[i]);
47         printf("\n");
48         for(int i = 0; i < tot; i++) printf("%d ", maxx[i]);
49         printf("\n");
50     }
51     return 0;
52 }

y @ The Angry Teletubbies 2011-11-05 09:54 发表评论

hdoj 4006 The kth great number - 优先队列

y @ The Angry Teletubbies — Sat, 17 Sep 2011 12:47:00 GMT

【题意】：(x��)�l�出一个n和k�Q�表�C�有n个操作。操作只有两�U�：(x��)1.插入一个数x 2.查询当前�W�k��的数�?br />
【题解】：(x��)题意很好理解�Q�这里可以直接用stl �?priority queue�Q�代码超�U�短。维护一个优先队列，佉K��列中元素个数��M��过k�Q�查询时直接输出堆顶元素卛_��?br />
【代码】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 #include "queue"
5 #include "functional"
6 #include "string"
7 using namespace std;
8 int n, k;
9 void solve() {
10     int val;
11     string s;
12     priority_queue<int, vector<int>, greater<int> > que;
13     for(int i = 0; i < n; i++) {
14         cin >> s;
15         if(s[0] == 'I') {
16             scanf("%d", &val);
17             que.push(val);
18             if(que.size() > k) que.pop();
19         } else cout << que.top() << endl;
20     }
21 }
22
23 int main() {
24     while(~scanf("%d%d", &n, &k)) solve();
25     return 0;
26 }