hdoj 4324 Triangle LOVE威尼斯人棋牌

hdoj 4324 Triangle LOVE

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 3903 Accepted Submission(s): 1537

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威尼斯人棋牌, = 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注意。注意,注意,取边,用字符串,不然超时!!!

#include #include #include using namespace std; #include #include #define N 2010 struct line { int u,v,next; }edge[N*N]; int n,m,inde[N],head[N],top=0; void add(int u,int v) { edge[top].u =u; edge[top].v =v; edge[top].next =head[u]; head[u]=top  ; inde[v]  ; } void topo() { int i,j,t=0,k=0,s; queueQ; for(i=0;i

 

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