Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3574 Accepted Submission(s): 1401
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”.
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).
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.
Sample Input
2 5 00100 10000 01001 11101 11000 5 01111 00000 01000 01100 01110
Sample Output
Case #1: Yes Case #2: No
#include<stdio.h> #include<string.h> char p[2001][2001]; int s[2001],m; void TO(int n) { int flag=0; int i,j,k; for(i=0;i<n;i++) { for(j=0;j<n;j++) if(s[j]==0) break; if(j==n) { flag=1; break; } s[j]=-1; for(k=0;k<n;k++) if(p[j][k]=='1'&&s[k]!=0) s[k]--; } printf("Case #%d: ",++m); if(flag) printf("Yes\n"); else printf("No\n"); } int main() { int t,n; m=0; scanf("%d",&t); while(t--) { memset(s,0,sizeof(s)); scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%s",p[i]); for(int j=0;j<n;j++) { if(p[i][j]=='1') s[j]++; } } TO(n); } return 0; }
再贴一个代码:
#include<stdio.h> #include<string.h> char p[2001][2001]; int s[2001],m; void TO(int n) { int flag=0,i,j,k,w=1000000,h; for(i=0;i<n;i++) { for(j=0;j<n;j++) if(s[j]==0) { w=j;break; } else w=1000000; if(w==1000000) { printf("Case #%d: ",++m); printf("Yes\n"); return ; } s[w]=-1; for(k=0;k<n;k++) if(p[w][k]=='1') s[k]--; } printf("Case #%d: ",++m); printf("No\n"); return ; } int main() { int t,n; m=0; scanf("%d",&t); while(t--) { memset(s,0,sizeof(s)); scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%s",p[i]); for(int j=0;j<n;j++) { if(p[i][j]=='1') s[j]++; } } TO(n); } return 0; }
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