hdu3342，拓扑排序模板题

Legal or Not

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9655    Accepted Submission(s): 4489

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 20 11 22 20 11 00 0

 

Sample Output

YES

NO

题目大意：n个人中输入m种关系，问是否有冲突的，比如a是b的领导，b是c的领导，然后输入c是a的领导这就矛盾了，形成了环式关系图，可以想到用拓扑排序来判环~~刚开始用了并查集，似乎不太合适，并查集更适合朋友之间的关系，问一群人是不是朋友，这种其实更适合并查集。

c++代码如下：

#include<stdio.h>
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m)&&n)
    {
        int a[101][101]={0},x,y,ru[101]={0};
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&x,&y);
           if(!a[x][y])//这里很坑~~评测数据会有重复的，卡了好几次
           {
                a[x][y]=1;//y->x,y的上级是x
            ru[x]++;//x入度加1
           }


        }
        int sum=0,vis[100]={0};
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(ru[j]==0&&vis[j]==0)
                {sum++;
                vis[j]=1;
                    for(int k=0;k<n;k++)
                    {
                        if(a[k][j]==1)
                        {
                            a[k][j]=0;
                            ru[k]--;
                        }
                    }
                }

            }
        }
        if(sum==n)printf("YES\n");
        else printf("NO\n");

    }
}

这里展示一下并查集写法，但是超内存，请网友们批评改正

#include
#include
#include
#include
using namespace std;
int pre[103];
a[100][100];
int uu=0;
int find_(int t,int y)
{
    if(pre[t]==t)
    {
        return t;
    }
    int o=pre[t];
    pre[t]=find_(pre[t],y);
    if(a[t][pre[t]]==1)
    {
        uu=1;
    }
    else
    { a[pre[t]][y]=1;
        a[o][y]=1;
    }

    return pre[t];
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m)&&n&&m)
    {
        int x,y;
        uu=0;
        for(int i=0;i<n;i++)
            pre[i]=i;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&x,&y);
            int xx=find_(x,y);
            if(a[y][x]==1)uu=1;
            else
            {
                 a[x][y]=1;
                 pre[y]=x;
            }
    }

        if(uu==1)printf("NO\n");
            else printf("YES\n");
}
}