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poj 1743 Musical Theme - 求不可重叠最镉K��复子�?+ height数组分组 + 二分

y @ The Angry Teletubbies — Mon, 14 May 2012 04:31:00 GMT

【题意】：(x��)求一个字�W�串的不可重叠最镉K��复子丌Ӏ?br />
【题解】：(x��)后缀数组�l�典题目�?br /> 先二分答案，把题目变成判定性问题：(x��)判断是否存在两个长度为k 的子串是相同的，且不重叠。解册��个问题的关键是利用height 数组�Q�把排序后的后缀分成若干�l�，其中每组的后�~�之间的height值都不小于k�?br /> �Ҏ(gu��)��看出�Q�有希望成�ؓ(f��)最长公共前�~�不小于k 的两个后�~�一定在同一�l�。然后对于每�l�后�~��Q�只��d��断每个后�~�的sa 值的最大值和最��g��差是否不��于k。如果有一�l�满��I��则说明存在，否则不存在。时间复杂度为O(nlogn)�?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 #include "set"
13 #include "utility"
14 using namespace std;
15 typedef pair<int, int> pii;
16 #define pb push_back
17 #define mp make_pair
18 #define fi first
19 #define se second
20 #define sof(x) sizeof(x)
21 #define lc(x) (x << 1)
22 #define rc(x) (x << 1 | 1)
23 #define lowbit(x) (x & (-x))
24 #define ll long long
25 #define maxn 20050
26 int wa[maxn], wb[maxn], wv[maxn], wc[maxn];
27 int r[maxn], sa[maxn], rank[maxn], height[maxn];
28 int n;
29
30 int cmp(int *r, int a, int b, int l) {
31     return r[a] == r[b] && r[a+l] == r[b+l];
32 }
33
34 void da() {
35     //m为最大字�W?/span>
36     int i, j, p, *x = wa, *y = wb, *t, m = 256;
37     for(i = 0; i < m; i++) wc[i] = 0;
38     for(i = 0; i <= n; i++) wc[x[i] = r[i]]++;
39     for(i = 1; i < m; i++) wc[i] += wc[i-1];
40     for(i = n; i >= 0; i--) sa[--wc[x[i]]] = i;
41     for(j = 1, p = 1; p < n; j *= 2, m = p) {
42         for(p = 0, i = n - j + 1; i <= n; i++) y[p++] = i;
43         for(i = 0; i <= n; i++) if(sa[i] >= j) y[p++] = sa[i] - j;
44         for(i = 0; i <= n; i++) wv[i] = x[y[i]];
45         for(i = 0; i < m; i++) wc[i] = 0;
46         for(i = 0; i <= n; i++) wc[wv[i]]++;
47         for(i = 1; i < m; i++) wc[i] += wc[i-1];
48         for(i = n; i >= 0; i--) sa[--wc[wv[i]]] = y[i];
49         for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i <= n; i++)
50             x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p - 1 : p++;
51     }
52 }
53
54 void calheight() {
55     int i, j, k = 0;
56     for(i = 1; i <= n; i++) rank[sa[i]] = i;
57     for(i = 0; i < n; height[rank[i++]] = k)
58         for(k ? k-- : 0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
59 }
60
61 bool check(int mid) {
62     int minn = sa[1], maxx = sa[1];
63     for(int i = 2; i <= n; i++) {
64         if(height[i] >= mid) {
65             minn = min(minn, sa[i]);
66             maxx = max(maxx, sa[i]);
67             if(maxx - minn >= mid) return 1;
68         } else minn = maxx = sa[i];
69     }
70     return false;
71 }
72
73 void solve() {
74     da();//求sa数组
75     calheight();//求rank数组和height数组
76     int l = 1, r = n, ans = -1;
77     while(l <= r) {
78         int mid = (l + r) >> 1;
79         if(check(mid)) ans = mid, l = mid + 1;
80         else r = mid - 1;
81     }
82     if(ans >= 4) printf("%d\n", ans + 1);
83     else printf("0\n");
84 }
85
86 int main() {
87     int a, b;
88     while(~scanf("%d", &n)) {
89         if(!n) break;
90         n--;
91         scanf("%d", &b);
92         for(int i = 0; i < n; i++) {
93             scanf("%d", &a);
94             r[i] = a - b + 100;
95             b = a;
96         }
97         r[n] = 0;
98         solve();
99     }
100     return 0;
101 }
102

y @ The Angry Teletubbies 2012-05-14 12:31 发表评论

poj 2406 Power Strings - kmp + 求重复子串的长度

y @ The Angry Teletubbies — Mon, 20 Feb 2012 10:28:00 GMT

【题意】：(x��)�l�出一个串�Q�求出它重复子串的长�?br />
【题解】：(x��)kmp�Q�求出原串的next[]�?br /> 如果len % (len - next[len]) == 0�Q�说明原串是�׃��个子串连�l�拼接而成的，且重复子串的长度�?(len / (len - next[len]));

【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 1000050
19 char s[maxn];
20 int next[maxn];
21 void getnext() {
22     int i = 0, j = -1;
23     next[i] = j;
24     int len = strlen(s);
25     while(i < len) {
26         if(j == -1 || s[i] == s[j]) i++, j++, next[i] = j;
27         else j = next[j];
28     }
29 }
30
31 int main() {
32     while(~scanf("%s", s)) {
33         if(strcmp(s, ".") == 0) break;
34         getnext();
35         int len = strlen(s);
36         if(len % (len - next[len]) == 0) printf("%d\n", len / (len - next[len]));
37         else printf("1\n");
38     }
39     return 0;
40 }
41

y @ The Angry Teletubbies 2012-02-20 18:28 发表评论

y @ The Angry Teletubbies — Sat, 18 Feb 2012 15:54:00 GMT

【题意】：(x��)求串a在串b出现的次数�?br />
【题解】：(x��)kmp卛_��?br /> 我利用了后缀数组的思想�Q�构造新串A + '$' + B�Q�然后对��C��求next[]卛_��Q�具体看代码�?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 1050000
19 char s[maxn];
20 int slen, len;
21 int next[maxn];
22 int getnext() {
23     int i = 1, j = 0, cnt = 0;
24     next[1] = 0;
25     while(i <= len) {
26         if(j == 0 || s[i] == s[j]) {
27             i++, j++, next[i] = j;
28             if(next[i] == slen + 1) cnt++;
29         } else j = next[j];
30     }
31     return cnt;
32 }
33
34 int main() {
35     int T;
36     scanf("%d", &T);
37     while(T--) {
38         scanf("%s", s + 1);
39         slen = strlen(s + 1);
40         s[slen+1] = '$';
41         scanf("%s", s + slen + 2);
42         len = strlen(s + 1);
43         printf("%d\n", getnext());
44     }
45     return 0;
46 }
47

y @ The Angry Teletubbies 2012-02-18 23:54 发表评论

hdoj 2243 考研路茫茫——单词情�l?- ac自动�?+ 矩阵快速幂 + 快速幂 + 二分

y @ The Angry Teletubbies — Mon, 17 Oct 2011 09:24:00 GMT

摘要: 【题意】：(x��)�l�定n个词根，求长度不��过L�Q�至��包含一个词根的字符串的数目。【题解】：(x��)可以从对立角度来惻I��是先求出所有字�W�串的数目，再减��M��包含��M��词根的字�W�串的数目�?nbsp; 字符串��L��?6^1 + 26^2 + …... 阅读全文

y @ The Angry Teletubbies 2011-10-17 17:24 发表评论

poj 2778 DNA Sequence - ac自动�?+ 矩阵快速幂 + 转化为求L步长的方案数

y @ The Angry Teletubbies — Mon, 17 Oct 2011 09:04:00 GMT

【题意】：(x��)�l�定你m个病毒串�Q�求长度为n的不包含��L��一个病毒串的种数�?br />
【题解】：(x��)用ac自动机构造一个DFA�Q�既有确定性的状态�{�U�d��Q�，然后用矩阵快速幂求解�?br />
【代码】：(x��)

  1 #include "iostream"
  2 #include "cstdio"
  3 #include "cstring"
  4 using namespace std;
  5 #define ll long long
  6 const ll MOD = 100000;
  7 const int kind = 4;
  8 const int maxn = 500;
  9 #define MAX 105
10 int root, tot;
11 int n, m;
12 int que[maxn], head, tail;
13 bool visit[maxn];
14 struct Node {
15     int child[kind];
16     int fail;
17     int end;
18     void init() {
19         memset(child, 0, sizeof(child));
20         fail = -1, end = 0;
21     }
22 } T[maxn];
23
24 void init() {
25     root = tot = 0;
26     T[root].init();
27 }
28
29 int hash(char ch) {
30     if(ch == 'A') return 0;
31     else if(ch == 'C') return 1;
32     else if(ch == 'G') return 2;
33     else return 3;
34 }
35
36 void insert(char *s) {//插入单词
37     int p = root, index;
38     while (*s) {
39         index = hash(*s);
40         if (!T[p].child[index]) {
41             T[++tot].init();
42             T[p].child[index] = tot;
43         }
44         p = T[p].child[index];
45         s++;
46     }
47     T[p].end = 1;
48 }
49
50 void build_ac_auto() {
51     head = tail = 0;
52     que[tail++] = root;
53     while (head < tail) {
54         int u = que[head++];
55         for (int i = 0; i < kind; i++) {
56             if (T[u].child[i]) {
57                 int son = T[u].child[i];
58                 int p = T[u].fail;
59                 if (u == root) T[son].fail = root;
60                 else {
61                     T[son].fail = T[p].child[i];
62                     T[son].end |= T[T[son].fail].end;
63                 }
64                 que[tail++] = son;
65             } else {//trie图，讑֮�虚拟节点
66                 int p = T[u].fail;
67                 if (u == root) T[u].child[i] = root;
68                 else T[u].child[i] = T[p].child[i];
69             }
70         }
71     }
72 }
73
74 struct Mat {
75     ll val[MAX][MAX];
76     void unit() {
77         zero();
78         for(int i = 0; i < MAX; i++) val[i][i] = 1;
79     }
80     void zero() {
81         memset(val, 0, sizeof(val));
82     }
83 }x;
84
85 Mat operator *(const Mat &a, const Mat &b) {
86     Mat tmp;
87     tmp.zero();
88     for(int k = 0; k <= tot; k++) {
89         for(int i = 0; i <= tot; i++) {
90             if(a.val[i][k])
91                 for(int j = 0; j <= tot; j++) {
92                     tmp.val[i][j] += a.val[i][k] * b.val[k][j];
93                     if(tmp.val[i][j] >= MOD) tmp.val[i][j] %= MOD;
94                 }
95         }
96     }
97     return tmp;
98 }
99
100 Mat operator ^(Mat x, int n) {
101     Mat tmp;
102     tmp.unit();
103     while(n) {
104         if(n & 1) tmp = tmp * x;
105         x = x * x;
106         n >>= 1;
107     }
108     return tmp;
109 }
110
111 void dfs(int u) {
112     visit[u] = true;
113     for(int i = 0; i < kind; i++) {
114         int v = T[u].child[i];
115         if(!T[v].end) {
116             x.val[u][v]++;
117             if(!visit[v]) dfs(v);
118         }
119     }
120 }
121
122 int main() {
123     char s[15];
124     while(~scanf("%d%d", &m, &n)) {
125         init();
126         for(int i = 0; i < m; i++) {
127             scanf("%s", s);
128             insert(s);
129         }
130         build_ac_auto();
131         x.zero();
132         memset(visit, false, sizeof(visit));
133         dfs(root);
134         Mat ans = x ^ n;
135         ll res = 0;
136         for(int i = 0; i <= tot; i++)
137             res += ans.val[0][i];
138         printf("%lld\n", res % MOD);
139     }
140     return 0;
141 }

y @ The Angry Teletubbies 2011-10-17 17:04 发表评论

zoj 3545 Rescue the Rabbit - AC自动�?+ 状态压�~�Dp

y @ The Angry Teletubbies — Mon, 03 Oct 2011 09:55:00 GMT

【题意】：(x��)�l�出n(n<11)个基因串�Q�每个基因串有对应的权��|��构造一个长度�ؓ(f��)l的基因，问最大的权��gؓ(f��)多少�Q�重复的基因只能��一�ơ�?br />
【题解】：(x��)考虑到最多只�?0个基因串�Q�我们可以��用状态压�~�。对于串之间的关�p�，我们可以建立对应的ac自动机，然后在ac自动��Z��q�行状态压�~�dp卛_��?br />              dp[i][j][k] i表示长度�Q�j表示基因�l�尾状态，k表示所含基因的状�?br />              转移方程�Q�设son为j的孩子节�?
              if(dp[i][j][k]) dp[i+1][son][k|T[son].id] = true;
              实现的时候我用了滚动数组压羃�I�间�Q�具体操作请看代码�?br />
【代码】：(x��)

  1 #include "iostream"
  2 #include "cstdio"
  3 #include "cstring"
  4 using namespace std;
  5 const int kind = 4;
  6 const int maxn = 250;
  7 int score[15];
  8 int root, tot;
  9 char ch[25];
10 int n, m, l;
11 int que[maxn], head, tail;
12
13 struct Node {
14     int child[kind];
15     int fail;
16     int id;
17     void init() {
18         memset(child, 0, sizeof (child));
19         fail = -1, id = 0;
20     }
21 } T[maxn];
22
23 void init() {
24     root = tot = 0;
25     T[root].init();
26 }
27
28 int hash(char ch) {
29     if(ch == 'A') return 0;
30     else if(ch == 'C') return 1;
31     else if(ch == 'G') return 2;
32     else return 3;
33 }
34
35 void insert(char *s, int id) {//插入单词
36     int p = root, index;
37     while (*s) {
38         index = hash(*s);
39         if (!T[p].child[index]) {
40             T[++tot].init();
41             T[p].child[index] = tot;
42         }
43         p = T[p].child[index];
44         s++;
45     }
46     T[p].id |= 1 << id;
47 }
48
49 void build_ac_auto() {
50     head = tail = 0;
51     que[tail++] = root;
52     while (head < tail) {
53         int u = que[head++];
54         for (int i = 0; i < kind; i++) {
55             if (T[u].child[i]) {
56                 int son = T[u].child[i];
57                 int p = T[u].fail;
58                 if (u == root) T[son].fail = root;
59                 else {
60                     T[son].fail = T[p].child[i];
61                     T[son].id |= T[T[son].fail].id;
62                 }
63                 que[tail++] = son;
64             } else {//trie图，讑֮�虚拟节点
65                 int p = T[u].fail;
66                 if (u == root) T[u].child[i] = root;
67                 else T[u].child[i] = T[p].child[i];
68             }
69         }
70     }
71 }
72 bool dp[2][205][1<<10];//滚动数组,不用滚动也可�?/span>
73 void solve() {
74     int mask = 1 << n, now, next;
75     memset(dp, false, sizeof(dp));
76     dp[0][0][0] = true;
77     for(int i = 0; i < l; i++) {
78         now = i % 2;
79         next = (i + 1) % 2;
80         memset(dp[next], false, sizeof(dp[next]));
81         for(int j = 0; j <= tot; j++) {
82             for(int k = 0; k < mask; k++) {
83                 if(dp[now][j][k]) {
84                     for(int p = 0; p < kind; p++) {
85                         int son = T[j].child[p];
86                         dp[next][son][k|T[son].id] = true;
87                     }
88                 }
89             }
90         }
91     }
92     int ans = -(1 << 30);
93     now = l % 2;
94     for(int i = 0; i <= tot; i++) {
95         for(int j = 0; j < mask; j++) {
96             if(dp[now][i][j]) {
97                 int t = 0;
98                 for(int k = 0; k < n; k++) {
99                     if(j & (1<< k)) t += score[k];
100                 }
101                 ans = max(ans, t);
102             }
103         }
104     }
105     if(ans < 0) printf("No Rabbit after 2012!\n");
106     else printf("%d\n", ans);
107 }
108
109 int main() {
110     while(~scanf("%d%d", &n, &l)) {
111         init();
112         for(int i = 0; i < n; i++) {
113             scanf("%s%d", ch, &score[i]);
114             insert(ch, i);
115         }
116         build_ac_auto();
117         solve();
118     }
119     return 0;
120 }

y @ The Angry Teletubbies 2011-10-03 17:55 发表评论

y @ The Angry Teletubbies — Wed, 21 Sep 2011 14:50:00 GMT

【题意】：(x��)�l�出n个单词，和一个文本，单词可能�?x��)重复出玎ͼ�问文本中一共出��C��多少个给定的单词�Q?br />
【题解】：(x��)AC自动机模杉K��?br />
【代码】：(x��)

1 #include "iostream"
2 #include "cstdio"
3 #include "cstring"
4 using namespace std;
5 #define maxn 700050
6 const int kind = 26;
7 int root, tot;
8 int que[maxn], head, tail;
9 int n, m;
10 char t[1005000];
11
12 struct Node {
13     int child[kind], fail, cnt;
14     void init() {
15         memset(child, 0, sizeof (child));
16         fail = -1, cnt = 0;
17     }
18 }T[maxn];
19
20 void init() {
21     root = tot = 0;
22     T[root].init();
23 }
24
25 void insert(char *s) {//插入单词
26     int p = root, index;
27     while (*s) {
28         index = *s - 'a';
29         if (!T[p].child[index]) {
30             T[++tot].init();
31             T[p].child[index] = tot;
32         }
33         p = T[p].child[index];
34         s++;
35     }
36     T[p].cnt++;
37 }
38
39 void build_ac_auto() {
40     head = tail = 0;
41     que[tail++] = root;
42     while (head < tail) {
43         int u = que[head++];
44         for (int i = 0; i < kind; i++) {
45             if (T[u].child[i]) {
46                 int son = T[u].child[i];
47                 int p = T[u].fail;
48                 if (u == root) T[son].fail = root;
49                 else T[son].fail = T[p].child[i];
50                 que[tail++] = son;
51             } else {//trie图，讑֮�虚拟节点
52                 int p = T[u].fail;
53                 if (u == root) T[u].child[i] = root;
54                 else T[u].child[i] = T[p].child[i];
55             }
56         }
57     }
58 }
59
60 int query(char *t) {
61     int p = root, cnt = 0;
62     while(*t) {
63         int index = *t - 'a';
64         p = T[p].child[index];
65         int tmp = p;
66         while(tmp != root) {
67             cnt += T[tmp].cnt;
68             T[tmp].cnt = 0;
69             tmp = T[tmp].fail;
70         }
71         t++;
72     }
73     return cnt;
74 }
75
76 int main() {
77     int T;
78     char s[55];
79     scanf("%d", &T);
80     while(T--) {
81         scanf("%d", &n);
82         init();
83         for(int i = 0; i < n; i++) {
84             scanf("%s", s);
85             insert(s);
86         }
87         scanf("%s", t);
88         build_ac_auto();
89         int ans = query(t);
90         printf("%d\n", ans);
91     }
92     return 0;
93 }

y @ The Angry Teletubbies 2011-09-21 22:50 发表评论