Triangle LOVE(HDU-4324)
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
01110Sample Output
Case #1: Yes
Case #2: No
题意:给出一任意两点间有且仅有一条单向边的有向图,要求判断这个图中是否存在仅有3个结点构成的环
思路:本质是一拓扑排序题
如果存在三个结点的环,则不能进行拓扑排序,
假设有一个 a->b->c->d ... 构成的环,由于任意两点之间一定存在边,对于 a、c 来说,若边为 c->a 则构成一个 3 结点的环,否则 a->c->d-> ... 构成一个 n-1 结点的环,以此类推,必然有 n-1、n-2、...、3 结点的环,因此,若能拓扑排序,则必定存在三节点的环
故对图进行拓扑排序判断即可
Source Program
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-6
#define MOD 16007
#define INF 0x3f3f3f3f
#define N 10001
#define LL long long
using namespace std;
int n,m;
int in[N];//节点入度
char str[N];
vector<int> G[N];//G[i]表示i节点所指向的所有其他点
bool judgeTopsort()//判断该图是否可拓扑排序
{
stack<int> S;
int cnt=0;//记录可拆解的点数目
for(int i=0;i<n;i++)//枚举编号从1到n的点
if(in[i]==0)//入度为0,入栈
S.push(i);
while(!S.empty()) {
int x=S.top();//取栈顶元素
S.pop();
cnt++;//可拆点数+1
for(int i=0;i<G[x].size();i++){
int y=G[x][i];
in[y]--;//入度减一
if(in[y]==0)//入度为0,出栈
S.push(y);
}
}
if(cnt==n)//AOV网点数等于图的点数,不存在环,可进行拓扑排序
return true;
else//AOV网点数等于图的点数,存在环,不可进行拓扑排序
return false;
}
int main()
{
int t;
int Case=1;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(in,0,sizeof(in));
for(int i=0;i<n;i++){
G[i].clear();
scanf("%s",str);
for(int j=0;j<n;j++){
if(str[j]=='1'){
G[i].push_back(j);
in[j]++;
}
}
}
printf("Case #%d: %s\n",Case++,judgeTopsort()?"No":"Yes");
}
return 0;
}