hdu4324(拓扑排序)
Triangle LOVE
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 4238 Accepted Submission(s): 1669
Problem Description
Recently, scientists find that there is love between any of two people. For example, between A and B, if A don’t love B, then B must love A, vice versa. And there is no possibility that two people love each other, what a crazy world!
Now, scientists want to know whether or not there is a “Triangle Love” among N people. “Triangle Love” means that among any three people (A,B and C) , A loves B, B loves C and C loves A.
Your problem is writing a program to read the relationship among N people firstly, and return whether or not there is a “Triangle Love”.
Now, scientists want to know whether or not there is a “Triangle Love” among N people. “Triangle Love” means that among any three people (A,B and C) , A loves B, B loves C and C loves A.
Your problem is writing a program to read the relationship among N people firstly, and return whether or not there is a “Triangle Love”.
Input
The first line contains a single integer t (1 <= t <= 15), the number of test cases.
For each case, the first line contains one integer N (0 < N <= 2000).
In the next N lines contain the adjacency matrix A of the relationship (without spaces). Ai,j = 1 means i-th people loves j-th people, otherwise Ai,j = 0.
It is guaranteed that the given relationship is a tournament, that is, Ai,i= 0, Ai,j ≠ Aj,i(1<=i, j<=n,i≠j).
For each case, the first line contains one integer N (0 < N <= 2000).
In the next N lines contain the adjacency matrix A of the relationship (without spaces). Ai,j = 1 means i-th people loves j-th people, otherwise Ai,j = 0.
It is guaranteed that the given relationship is a tournament, that is, Ai,i= 0, Ai,j ≠ Aj,i(1<=i, j<=n,i≠j).
Output
For each case, output the case number as shown and then print “Yes”, if there is a “Triangle Love” among these N people, otherwise print “No”.
Take the sample output for more details.
Take the sample output for more details.
Sample Input
2 5 00100 10000 01001 11101 11000 5 01111 00000 01000 01100 01110
Sample Output
Case #1: Yes Case #2: No
标准的拓扑排序。。。。可恶的是这题卡输入,真的是无语死,用scanf("%s",)就能过,scanf("%c")就超时。。。。。真涨姿势
#include <iostream> #include <stdio.h> #include <stdlib.h> #include<string.h> #include<algorithm> #include<math.h> #include<queue> using namespace std; typedef long long ll; int o=1; const int maxn=2010; int head[maxn],ip,rd[maxn]; struct data { int v,next; } tu[maxn*maxn]; void init(int n) { ip=0; memset(rd,0,sizeof(rd)); memset(head,-1,sizeof(head)); } void add(int u,int v) { tu[ip].v=v,tu[ip].next=head[u],head[u]=ip++; } void topu(int n) { int s=0; queue<int>q; for(int i=0; i<n; i++) if(!rd[i]) q.push(i); while(!q.empty()) { int tem=q.front(); q.pop(); s++; for(int i=head[tem]; i!=-1; i=tu[i].next) { int hh=tu[i].v; rd[hh]--; if(!rd[hh]) q.push(hh); } } if(s==n) printf("Case #%d: No\n",o++); else printf("Case #%d: Yes\n",o++); } char x[2010]; int main() { int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); init(n); for(int i=0; i<n; i++) { scanf("%s",x); for(int j=0; j<n; j++) { if(x[j]=='1') add(i,j),rd[j]++; } } topu(n); } return 0; }
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