1 /* 鎷撴墤鎺掑簭 */
2
3 #include<iostream>
4 #include<vector>
5 using namespace std;
6 const int MAXN=120;
7 vector<int> adj[MAXN];
8
9 int into[MAXN];
10
11 int main()
12 {
13 //freopen("data.in","r",stdin);
14 //freopen("data.out","w",stdout);
15 int n;
16 memset(into,0,sizeof(into));
17 cin >> n;
18 for (int i=1;i<=n;++i)
19 {
20 int x;
21 cin >> x;
22 while (x!=0)
23 {
24 adj[i].push_back(x);
25 ++into[x];
26 cin >> x;
27 }
28 }
29 //for (int i=1;i<=n;++i) cout << into[i] << ' '; cout << endl;
30 bool everp=0;
31 for (int k=0;k<n;++k)
32 {
33 int i=1;
34 for (;i<=n;i++) if (!into[i]) break;
35 (everp) ? cout << ' ' << i : cout << i;
36 everp=1;
37 --into[i];
38 for (int j=0;j<adj[i].size();++j) --into[ adj[i][j] ];
39 }
40 cout << endl;
41 return 0;
42 }
43
2
3 #include<iostream>
4 #include<vector>
5 using namespace std;
6 const int MAXN=120;
7 vector<int> adj[MAXN];
8
9 int into[MAXN];
10
11 int main()
12 {
13 //freopen("data.in","r",stdin);
14 //freopen("data.out","w",stdout);
15 int n;
16 memset(into,0,sizeof(into));
17 cin >> n;
18 for (int i=1;i<=n;++i)
19 {
20 int x;
21 cin >> x;
22 while (x!=0)
23 {
24 adj[i].push_back(x);
25 ++into[x];
26 cin >> x;
27 }
28 }
29 //for (int i=1;i<=n;++i) cout << into[i] << ' '; cout << endl;
30 bool everp=0;
31 for (int k=0;k<n;++k)
32 {
33 int i=1;
34 for (;i<=n;i++) if (!into[i]) break;
35 (everp) ? cout << ' ' << i : cout << i;
36 everp=1;
37 --into[i];
38 for (int j=0;j<adj[i].size();++j) --into[ adj[i][j] ];
39 }
40 cout << endl;
41 return 0;
42 }
43
Chen Jiecao 2009-07-21 17:14 鍙戣〃璇勮
]]> 1 #include<iostream>
2 using namespace std;
3 const int MAXN=50005;
4
5
6 int main()
7 {
8 freopen("data.in","r",stdin);
9 freopen("data.out","w",stdout);
10 int c,cc,num[MAXN];
11 cin >> c;
12 for (int i=0;i<c;++i) cin >> num[i];
13 num[c]=31440461;
14 cin >> cc;
15 int p=0;
16 for (int i=0;i<cc;++i)
17 {
18 int x;
19 cin >> x;
20 while (x+num[p]<10000) ++p;
21 if (x+num[p]==10000) { cout << "YES\n"; return 0; }
22 }
23 cout << "NO\n";
24 return 0;
25 }
26
2 using namespace std;
3 const int MAXN=50005;
4
5
6 int main()
7 {
8 freopen("data.in","r",stdin);
9 freopen("data.out","w",stdout);
10 int c,cc,num[MAXN];
11 cin >> c;
12 for (int i=0;i<c;++i) cin >> num[i];
13 num[c]=31440461;
14 cin >> cc;
15 int p=0;
16 for (int i=0;i<cc;++i)
17 {
18 int x;
19 cin >> x;
20 while (x+num[p]<10000) ++p;
21 if (x+num[p]==10000) { cout << "YES\n"; return 0; }
22 }
23 cout << "NO\n";
24 return 0;
25 }
26
Chen Jiecao 2009-07-21 17:12 鍙戣〃璇勮
]]> 1 #include<iostream>
2 #include<math.h>
3 using namespace std;
4 const double PI=3.1415926;
5
6 template <class T>
7 T dis(T x1,T y1,T x2,T y2)
8 {
9 return sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) );
10 }
11 int main()
12 {
13 int n;
14 double r,len=0;
15 //freopen("data.in","r",stdin);
16 //freopen("data.out","w",stdout);
17 cin >> n >> r;
18 double x0,y0,x,y;
19 cin >> x0 >> y0;
20 double px=x0,py=y0;
21 if (n==1) goto L1;
22 for (int i=1;i<n;++i)
23 {
24 cin >> x >> y;
25 len+=dis(px,py,x,y);
26 px=x;
27 py=y;
28 }
29 len+=dis(x,y,x0,y0);
30 L1: len+= 2*PI*r ;
31 printf("%.2f\n",len);
32 return 0;
33 }
34
2 #include<math.h>
3 using namespace std;
4 const double PI=3.1415926;
5
6 template <class T>
7 T dis(T x1,T y1,T x2,T y2)
8 {
9 return sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) );
10 }
11 int main()
12 {
13 int n;
14 double r,len=0;
15 //freopen("data.in","r",stdin);
16 //freopen("data.out","w",stdout);
17 cin >> n >> r;
18 double x0,y0,x,y;
19 cin >> x0 >> y0;
20 double px=x0,py=y0;
21 if (n==1) goto L1;
22 for (int i=1;i<n;++i)
23 {
24 cin >> x >> y;
25 len+=dis(px,py,x,y);
26 px=x;
27 py=y;
28 }
29 len+=dis(x,y,x0,y0);
30 L1: len+= 2*PI*r ;
31 printf("%.2f\n",len);
32 return 0;
33 }
34
Chen Jiecao 2009-07-20 16:49 鍙戣〃璇勮
]]>Time Limit: 2.0 second
Memory Limit: 16 MB
Memory Limit: 16 MB
The segment of numerical axis from 0 to 109
is painted into white color. After that some parts of this segment are
painted into black, then some into white again and so on. In total
there have been made N re-paintings (1 ≤ N ≤ 5000). You are to write a program that finds the longest white open interval after this sequence of re-paintings.
Input
The first line of input contains the only number N. Next N lines contain information about re-paintings. Each of these lines has a form:
ai bi ci
where ai and bi are integers, ci is symbol 'b' or 'w', ai, bi, ci are separated by spaces.
This triple of parameters represents repainting of segment from ai to bi into color ci ('w' 鈥?white, 'b' 鈥?black). You may assume that 0 < ai < bi < 109.
ai bi ci
where ai and bi are integers, ci is symbol 'b' or 'w', ai, bi, ci are separated by spaces.
This triple of parameters represents repainting of segment from ai to bi into color ci ('w' 鈥?white, 'b' 鈥?black). You may assume that 0 < ai < bi < 109.
Output
Output should contain two numbers x and y (x < y)
divided by space(s). These numbers should define the longest white open
interval. If there are more than one such an interval output should
contain the one with the smallest x.
Sample
| input | output |
|---|---|
4 |
47 634 |
澶嶆潅搴︿笂鐣屼負O(M*M),榪欓噷M琛ㄧず紱繪暎鍖栧悗寰楀埌鐨勫尯闂翠釜鏁?
1 #include<iostream>
2 #include<algorithm>
3 using namespace std;
4 const int MAXN=10005;
5 const int MAXL=1000000000;
6 struct re
7 {
8 int a,b;
9 char c;
10 }op[MAXN];
11
12 int n,p=0,que[MAXN];
13 bool color[MAXN<<1];
14
15 /* 浜屽垎鏌ユ壘 */
16 int find(int num)
17 {
18 int l=0,r=p+1,mid;
19 while (r-l>1)
20 {
21 mid=(l+r) >> 1;
22 (que[mid]<=num)? l=mid : r=mid;
23 if (que[l]==num) return l;
24 }
25 return l;
26 }
27
28 void disp(re& op)
29 {
30 /* 鍖洪棿鐢?寮濮嬫暟,鑰屼笉鏄敱0寮濮?nbsp;*/
31 op.a=find(op.a)+1;
32 op.b=find(op.b);
33 }
34
35 void mark(int l,int r,char c)
36 {
37 for (int i=l;i<=r;++i) color[i]=(c=='b');
38 }
39
40 int main()
41 {
42 // freopen("data.in","r",stdin);
43 // freopen("data.out","w",stdout);
44 cin >> n;
45 for (int i=0;i<n;++i)
46 {
47 cin >> op[i].a >> op[i].b >> op[i].c;
48 que[++p]=op[i].a;
49 que[++p]=op[i].b;
50 }
51 que[++p]=MAXL;
52 /* 鍒犻櫎閲嶅鍑虹幇鐨勬暟瀛?nbsp;*/
53 int rp=0;
54 for (int i=1;i<=p;++i)
55 if (que[rp]!=que[i]) que[++rp]=que[i];
56 p=rp;
57 /* 鎺掑簭,渚夸簬瀹氫綅鍜岀鏁ｅ寲 */
58 sort(que,que+p+1);
59 que[p+1]=MAXL+1;
60 /* 瀵規(guī)搷浣滆繘琛岀鏁ｅ鐞?nbsp;*/
61 for (int i=0;i<n;++i) disp(op[i]);
62 /* 澶勭悊鎿嶄綔搴忓垪 */
63 memset(color,0,sizeof(color));
64 for (int i=0;i<n;++i) mark(op[i].a,op[i].b,op[i].c);
65 int t=1,a,b,mlen=0;
66 for (int i=2;i<=p+1;++i)
67 if (color[i]!=color[t] || i>p)
68 {
69 if ((!color[t]) && (que[i-1]-que[t-1]>mlen)) { a=que[t-1];b=que[i-1];mlen=que[i-1]-que[t-1];}
70 t=i;
71 }
72 cout << a << ' ' << b << endl;
73 //system("pause");
74 return 0;
75 }
76
涓埆鐩爣榪滃ぇ鐨勫悓瀛﹀彲鑳藉茍涓嶄粎浠呮槸鎯寵В鍐寵繖涓鐩?榪欐椂鍊欏緩璁綘浣跨敤紱繪暎鍖?綰挎鏍?澶嶆潅搴︿細澶уぇ鍦伴檷浣?br> 2 #include<algorithm>
3 using namespace std;
4 const int MAXN=10005;
5 const int MAXL=1000000000;
6 struct re
7 {
8 int a,b;
9 char c;
10 }op[MAXN];
11
12 int n,p=0,que[MAXN];
13 bool color[MAXN<<1];
14
15 /* 浜屽垎鏌ユ壘 */
16 int find(int num)
17 {
18 int l=0,r=p+1,mid;
19 while (r-l>1)
20 {
21 mid=(l+r) >> 1;
22 (que[mid]<=num)? l=mid : r=mid;
23 if (que[l]==num) return l;
24 }
25 return l;
26 }
27
28 void disp(re& op)
29 {
30 /* 鍖洪棿鐢?寮濮嬫暟,鑰屼笉鏄敱0寮濮?nbsp;*/
31 op.a=find(op.a)+1;
32 op.b=find(op.b);
33 }
34
35 void mark(int l,int r,char c)
36 {
37 for (int i=l;i<=r;++i) color[i]=(c=='b');
38 }
39
40 int main()
41 {
42 // freopen("data.in","r",stdin);
43 // freopen("data.out","w",stdout);
44 cin >> n;
45 for (int i=0;i<n;++i)
46 {
47 cin >> op[i].a >> op[i].b >> op[i].c;
48 que[++p]=op[i].a;
49 que[++p]=op[i].b;
50 }
51 que[++p]=MAXL;
52 /* 鍒犻櫎閲嶅鍑虹幇鐨勬暟瀛?nbsp;*/
53 int rp=0;
54 for (int i=1;i<=p;++i)
55 if (que[rp]!=que[i]) que[++rp]=que[i];
56 p=rp;
57 /* 鎺掑簭,渚夸簬瀹氫綅鍜岀鏁ｅ寲 */
58 sort(que,que+p+1);
59 que[p+1]=MAXL+1;
60 /* 瀵規(guī)搷浣滆繘琛岀鏁ｅ鐞?nbsp;*/
61 for (int i=0;i<n;++i) disp(op[i]);
62 /* 澶勭悊鎿嶄綔搴忓垪 */
63 memset(color,0,sizeof(color));
64 for (int i=0;i<n;++i) mark(op[i].a,op[i].b,op[i].c);
65 int t=1,a,b,mlen=0;
66 for (int i=2;i<=p+1;++i)
67 if (color[i]!=color[t] || i>p)
68 {
69 if ((!color[t]) && (que[i-1]-que[t-1]>mlen)) { a=que[t-1];b=que[i-1];mlen=que[i-1]-que[t-1];}
70 t=i;
71 }
72 cout << a << ' ' << b << endl;
73 //system("pause");
74 return 0;
75 }
76
Chen Jiecao 2009-07-20 15:00 鍙戣〃璇勮
]]>A Binary Apple Tree
Time Limit: 1.0 second
Memory Limit: 16 MB
Let's
imagine how apple tree looks in binary computer world. You're right, it
looks just like a binary tree, i.e. any biparous branch splits up to
exactly two new branches. We will enumerate by natural numbers the root
of binary apple tree, points of branching and the ends of twigs. This
way we may distinguish different branches by their ending points. We
will assume that root of tree always is numbered by 1 and all numbers
used for enumerating are numbered in range from 1 to N, where N is the total number of all enumerated points. For instance in the picture below N is equal to 5. Here is an example of an enumerated tree with four branches:
2 5 |
As
you may know it's not convenient to pick an apples from a tree when
there are too much of branches. That's why some of them should be
removed from a tree. But you are interested in removing branches in the
way of minimal loss of apples.
So your are given amounts of apples on a branches and amount of
branches that should be preserved. Your task is to determine how many
apples can remain on a tree after removing of excessive branches.
Input
First line of input contains two numbers: N and Q (1 ≤ Q ≤ N; 1 < N ≤ 100). N denotes the number of enumerated points in a tree. Q denotes amount of branches that should be preserved. Next N−1
lines contains descriptions of branches. Each description consists of a
three integer numbers divided by spaces. The first two of them define
branch by it's ending points. The third number defines the number of
apples on this branch. You may assume that no branch contains more than
30000 apples.
Output
Output should contain the only number 鈥?amount of apples that can be preserved. And don't forget to preserve tree's root ;-)
Sample
| input | output |
|---|---|
5 2 |
21 |
綆鏋?
榪欐槸涓涓畝鍗曠殑鏍戝艦鍔ㄦ佽鍒掗棶棰?澶ф鍙互鎷挎潵褰撹繖綾婚鐩殑鍏ラ棬璁粌棰?铏界劧榪欐槸URAL涓婄殑絎竴涓爲褰P,浣嗘槸鎴戝鎬殑鏄畠鐨勯氳繃鐜囧茍涓嶅緢楂?
瀵逛簬鍘熼鐩殑鍥懼艦,鐢ㄦ暟緇剉alue[a][b]琛ㄧずa,b鐐歸棿鑻規(guī)灉鐨勪釜鏁?鐢╪d[p].L,nd[p].R鍒嗗埆琛ㄧず鑺傜偣p鐨勫乏鍙沖効瀛?閫氳繃build_tree(1)鑾峰緱鏁扮粍nd[],浠庤岃幏寰楁暣媯墊爲鐨勪俊鎭?
鎺ョ潃,鐢╝ns[p][i]琛ㄧず浠ヨ妭鐐筽涓虹瀹楃殑瀛愭爲,淇濈暀鐨勬灊鏉′笉瓚呰繃i鏉℃椂鎵鑳戒繚鐣欑殑鏈澶氱殑鑻規(guī)灉,鐘舵佽漿縐繪湁涓涓嬪嚑縐嶆儏鍐?
鍦ㄩ櫎鍘誨浣欐灊鏉＄殑鍚庣殑鍥句腑,
1. p鍙笌涓涓効瀛愮浉榪?
ans[p][i]=max(ans[left_son][i-1]+value[left_son][p],ans[right_son][i-1]+value[right_son][p]);
2. p涓庝袱涓効瀛愮浉榪?
for (int j=0;j<=i-2;++j)
ans[p][i]=max(ans[p][i],ans[left_son][j]+ans[right_son][i-j-2]+d);
榪欓噷,d=value[left_son][p]+value[right_son][p];
綆楁硶鍦╫(N*Q*Q)綰у埆
1 #include<iostream>
2 using namespace std;
3 const int MAXN=102;
4 int n,q,value[MAXN][MAXN],ans[MAXN][MAXN];
5 struct node
6 {
7 int l,r;
8 }nd[MAXN];
9
10 void build_tree(int p)
11 {
12 int flg=0;
13 for (int i=1;i<=n;++i)
14 if (value[p][i] && (!nd[i].l))
15 {
16 flg=1;
17 if (nd[p].l==0) nd[p].l=i;
18 else
19 {nd[p].r=i; break;}
20 }
21 if (!flg) return;
22 if (nd[p].l) build_tree(nd[p].l);
23 if (nd[p].r) build_tree(nd[p].r);
24 }
25
26 void calc(int p)
27 {
28 if (!nd[p].l) return;
29 int l=nd[p].l,r=nd[p].r;
30 calc(l);
31 calc(r);
32 ans[p][1]=max(value[l][p],value[r][p]);
33
34 int d=value[l][p]+value[r][p];
35 for (int i=2;i<=q;++i)
36 {
37 ans[p][i]=max(ans[l][i-1]+value[l][p],ans[r][i-1]+value[r][p]);
38 for (int j=0;j<=i-2;++j)
39 ans[p][i]=max(ans[p][i],ans[l][j]+ans[r][i-j-2]+d);
40 }
41 }
42
43
44 int main()
45 {
46 //freopen("data.in","r",stdin);
47 //freopen("data.out","w",stdout);
48 cin >> n >> q;
49 memset(value,0,sizeof(value));
50 for (int i=1;i<n;++i)
51 {
52 int a,b,c;
53 cin >> a >> b >> c;
54 value[a][b]=c;
55 value[b][a]=c;
56 }
57 memset(nd,0,sizeof(nd));
58 build_tree(1);
59 calc(1);
60 cout << ans[1][q] << endl;
61 return 0;
62 }
63
2 using namespace std;
3 const int MAXN=102;
4 int n,q,value[MAXN][MAXN],ans[MAXN][MAXN];
5 struct node
6 {
7 int l,r;
8 }nd[MAXN];
9
10 void build_tree(int p)
11 {
12 int flg=0;
13 for (int i=1;i<=n;++i)
14 if (value[p][i] && (!nd[i].l))
15 {
16 flg=1;
17 if (nd[p].l==0) nd[p].l=i;
18 else
19 {nd[p].r=i; break;}
20 }
21 if (!flg) return;
22 if (nd[p].l) build_tree(nd[p].l);
23 if (nd[p].r) build_tree(nd[p].r);
24 }
25
26 void calc(int p)
27 {
28 if (!nd[p].l) return;
29 int l=nd[p].l,r=nd[p].r;
30 calc(l);
31 calc(r);
32 ans[p][1]=max(value[l][p],value[r][p]);
33
34 int d=value[l][p]+value[r][p];
35 for (int i=2;i<=q;++i)
36 {
37 ans[p][i]=max(ans[l][i-1]+value[l][p],ans[r][i-1]+value[r][p]);
38 for (int j=0;j<=i-2;++j)
39 ans[p][i]=max(ans[p][i],ans[l][j]+ans[r][i-j-2]+d);
40 }
41 }
42
43
44 int main()
45 {
46 //freopen("data.in","r",stdin);
47 //freopen("data.out","w",stdout);
48 cin >> n >> q;
49 memset(value,0,sizeof(value));
50 for (int i=1;i<n;++i)
51 {
52 int a,b,c;
53 cin >> a >> b >> c;
54 value[a][b]=c;
55 value[b][a]=c;
56 }
57 memset(nd,0,sizeof(nd));
58 build_tree(1);
59 calc(1);
60 cout << ans[1][q] << endl;
61 return 0;
62 }
63
Chen Jiecao 2009-07-19 23:02 鍙戣〃璇勮
]]>