hdu 3342 Legal or Not （判环问题）

Legal or Not

Problem Description

ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many "holy cows" like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost "master", and Lost will have a nice "prentice". By and by, there are many pairs of "master and prentice". But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?

We all know a master can have many prentices and a prentice may have a lot of masters too, it's legal. Nevertheless，some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian's master and, at the same time, 3xian is HH's master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not. 

Please note that the "master and prentice" relation is transitive. It means that if A is B's master ans B is C's master, then A is C's master.

 

Input

The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y's master and y is x's prentice. The input is terminated by N = 0.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.

 

Output

For each test case, print in one line the judgement of the messy relationship.
If it is legal, output "YES", otherwise "NO".

 

Sample Input

3 2
0 1
1 2
2 2
0 1
1 0
0 0

 

Sample Output

YES
NO

 

两种方法 ：

1，用拓扑的方法，每次去掉一个入度为0的点，全部点都去掉就是合法的。（下面代码用此法）

2，用深搜，记录路径，当搜到走过的路径上的点就是有环。每个点都可能重复搜多次，因为这不是树。

解题思路：

题目意思是：a打败b 那么a就是b的主人，如果b打败c 那么a和b都是c的主人，现在让你判断是否存在a打败b,b打败c,c打败a

这样不符合题意的情况出现，找到一个度为0的点，然后删除他可以到达的地方，然后一直删除，一直到没有停止，为了还有

点没删除那就说明存在环。

#include<stdio.h>
#include<iostream> 
#include <algorithm>
#include<string.h>
#include<vector>
#include<math.h>
#include<queue>
#include<set>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
int KGCD(int a,int b){if(a==0)return b;if(b==0)return a;if(~a&1){ if(b&1) return KGCD(a>>1,b);else return KGCD(a>>1,b>>1) <<1; } if(~b & 1)  return KGCD(a, b>>1);  if(a > b) return KGCD((a-b)>>1, b);return KGCD((b-a)>>1, a);}  
int LCM(int a,int b){ return a/KGCD(a,b)*b; } 
int dir[5][2]={0,1,0,-1,1,0,-1,0};
int map[110][110];//储存节点 
int in[110];    //储存入度情况 
int vis[110] ; //记录访问情况 
int main()  
{  
    int m,n;  
    while(scanf("%d%d",&n,&m)!=EOF)  
    {  
    	if(n==0)
    		break;
        memset(map,0,sizeof(map));  
        memset(vis,0,sizeof(vis));  
        memset(in,0,sizeof(in));  
        for(int i=0;i<m;i++)  
        {  
            int x,y;  
            scanf("%d%d",&x,&y);  
            if(map[x][y]==0)  
            {  
                map[x][y]=1; //标记这条边 
                in[y]++;    //标记入度 
            }  
        }  
        int count=0 ;  
        for(int i=0;i<n;i++)  
        {  
            for(int j=0;j<n;j++)  
            {  
                if(!vis[j]&&in[j]==0) //如果当前还没有访问过  而且没有出度 
                {  
                    count++;  
                    vis[j]=1;   
                    for(int k=0;k<n;k++)  
                    {  
                        if(map[j][k]==1)  //去掉该点到达的所有的点 
                        {  
                            map[j][k]=0;  
                            in[k]--;  
                        }  
                    }  
                    break; //重新扫描到一个入度为0的 
                }  
            }  
        }  
        if(count<n) //如果不是所有点都删除 那就说明有环 
        {  
            printf("NO\n");  
        }  
        else  
        {  
            printf("YES\n");  
        }  
    }  
      
    return 0 ;  
}