Legal or Not（bellman-ford）⭐⭐⭐

问题 G: Legal or Not

时间限制: 1 Sec  内存限制: 32 MB
提交: 32  解决: 10
[提交] [状态] [命题人:外部导入]

题目描述

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.

 

输入

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.

 

输出

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

 

样例输入

4 3
0 1
1 2
2 3
3 3
0 1
1 2
2 0
0 1

 

样例输出

YES
NO

一看见这个题目第一个思路是并查集，但是考虑到父节点的处理和判断，有点屡不清思路。

所以选择用这个算法来写，需要注意的是这个算法是用来检测单源负权回路的（局限还是蛮多的-_-||），但是好歹这个题可以用，_(:з」∠)_ 

那么，贴上代码。

 实现代码：

 

  #include<stdio.h>
    int main(void)
    {
        int dis[110],bak[110],i,k,n,m,u[110],v[110],w[110],check,flag;
        int inf=99999999;
        while(scanf("%d",&n),n!=0)
        {
            scanf("%d",&m);
         
            for(i=1;i<=m;i++)
            scanf("%d%d",&v[i],&u[i]);
             
        for(i=1;i<=m;i++)
        w[i]=-1;
         
        for(i=1;i<=n;i++)
           dis[i]=inf;
        dis[1]=0;
        for(k=1;k<=n-1;k++)
        {
            check=0;
        /*  for(i=1;i<=n;i++)
                bak[i]=dis[i];*/
            for(i=1;i<=m;i++)
               if(dis[v[i]]>dis[u[i]]+w[i])
               {
                    dis[v[i]]=dis[u[i]]+w[i];
                    check=1;
               }
                   
         
        /*  for(i=1;i<=n;i++)
               if(bak[i]!=dis[i])
               check=1;  */
            if(check==0)  break;
        }
        flag=0;
        for(i=1;i<=m;i++)
           if(dis[v[i]]>dis[u[i]]+w[i])
              flag=1;
             
             
    /*  for(i=1;i<=n;i++)
        printf("%d",dis[i]);*/
         
        if(flag==1)
           printf("NO\n");
        else
        printf("YES\n");
        }
        return 0;
    }

注意点的录入那个操作。