HDU 4324 Triangle LOVE 【拓扑排序】

Triangle LOVE

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3494    Accepted Submission(s): 1356

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). A_i,j = 1 means i-th people loves j-th people, otherwise A_i,j = 0.
It is guaranteed that the given relationship is a tournament, that is, A_i,i= 0, A_i,j ≠ A_j,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

25001001000001001111011100050111100000010000110001110

 

 

Sample Output

Case #1: YesCase #2: No

 

嗯，题目大意就是说，个人认为判断是否能成环吧，若能，则topsort查找到最后一定存在入度不为0 的点，若不能，则一定不存在入度不为0的点

 

 

#include <iostream>#include<cstdio>#include<cstring>#include<queue>#define maxn 2020using namespace std;int cnt,flag,head[maxn],in[maxn];queue<int>q;struct node{    int a,b,next;};node edge[maxn*maxn];void add(int a,int b){    edge[cnt].a=a;    edge[cnt].b=b;    edge[cnt].next=head[a];    head[a]=cnt;    cnt++;}void top_sort(){    int u,v;    while(!q.empty())    {        u=q.front();        q.pop();        for(int i=head[u];i!=-1;i=edge[i].next)        {            v=edge[i].b;            in[v]--;            if(in[v]==0)            {                in[v]=-1;                q.push(v);            }        }    }}int main(){    int t,n,num=0;    char s[maxn];    scanf("%d",&t);    while(t--)    {        num++;        cnt=flag=0;        memset(in,0,sizeof(in));        memset(head,-1,sizeof(head));        while(!q.empty())            q.pop();        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf("%s",s);            for(int j=0;j<n;++j)            {                if(s[j]=='1')                {                    add(i,j);                    in[j]++;                }            }        }        for(int i=0;i<n;++i)        {            if(in[i]==0)            {                in[i]=-1;                q.push(i);            }            top_sort();        }        for(int i=0;i<n;++i)        {            if(in[i]!=-1)            {                flag=1;                break;            }        }        printf("Case #%d: ",num);        if(flag)            printf("Yes\n");        else            printf("No\n");    }    return 0;}