hdu4324 Triangle Love【拓扑排序】有向无环图

 

/*
5
00100
10000
01001
11101
11000
5
01111
00000
01000
01100
01110
*/
#include<iostream>
#include<cstdio>
using namespace std;
#define N 2010
int indegree[N],a[N][N];
char b[N];
int main(){
//    freopen("in.txt","r",stdin);
    int n,t,x,p=1;
    scanf("%d",&t);
        while(t--){
            int sign1=0;
            scanf("%d",&n);
            for(int i=0;i<=n;i++){
                indegree[i]=0;
            }
            for(int i=1;i<=n;i++){
                scanf("%s",b);
                for(int j=0;j<n;j++){
                    a[i][j+1]=b[j]-'0';
                    if(a[i][j+1]==1)
                    indegree[j+1]++;//入度数组
                    cout<<i<<" "<<j+1<<" "<<indegree[j+1]<<endl;
                }
            }
            printf("Case #%d: ",p++);
///首先要找到任意入度为0的一个顶点，删除它及所有相邻的边，再找入度为0的顶点，
///以此类推，直到删除所有顶点。顶点的删除顺序即为拓扑排序。
            for(int i=1;i<=n;i++){
                int temp;
                int sign=0;
                for(int j=1;j<=n;j++){
                    if(indegree[j]==0)
                    {
                        temp=j;
                        indegree[j]=-1;
                        sign=1;
                        break;
                    }
                }
                if(sign==0){//如果没有找到定点，那一定有环了
                    printf("Yes\n");
                    sign1=1;
                    break;
                }
                else{
                    for(int j=1;j<=n;j++){
                        if(a[temp][j]&&temp!=j)
                            indegree[j]--;
                            cout<<"---:"<<temp<<" "<<j<<" "<<indegree[j]<<endl;

                    }
                }
            }
            if(sign1==0){
                printf("No\n");
            }
        }
    return 0;
}

不知道什么时候可以写自己的代码，这是学了数据结构以来第一次听到拓扑排序。应该很好用。继续保留！
Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 994    Accepted Submission(s): 407

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