hdu 5154 Harry and Magical Computer（拓扑排序）

Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1845 Accepted Submission(s): 727



Problem Description
In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with.
We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer
can finish all the n processes.


Input
There are several test cases, you should process to the end of file.

For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies.
1≤n≤100,1≤m≤10000

The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b).
1≤a,b≤n


Output
Output one line for each test case.

If the computer can finish all the process print "YES" (Without quotes).

Else print "NO" (Without quotes).


Sample Input

3 2
3 1
2 1
3 3
3 2
2 1
1 3

Sample Output

YES
NO

题意：一共有n个任务，共有m条限制条件，对于m条限制条件来说：要完成a，需要先完成b。问最后能否完成所有任务。

a依赖于b，要完成a，需要完成任务b。典型的拓扑排序条件：全序条件。这里直接上代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int a[110][110];
int degree[110];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(a,0,sizeof(a));
        memset(degree,0,sizeof(degree));
        for(int i=0;i<m;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            if(a[x][y]==0)
            {
                a[x][y]=1;
                degree[y]++;//入度
            }
        }
        int jieshu=0;
        int cont=0;
        while(1)
        {
            int j=1;
            if(degree[j]!=0)
            {
                while(1)
                {
                    j++;
                    if(degree[j]==0)break;
                    if(j>n)break;
                }
            }
            if(j>n)//如果说当前没有了度为0的点，break就行。
            break;
            degree[j]=-1;
            cont++;
            if(cont==n)
            {
                jieshu=1;
                break;
            }
            for(int i=1;i<=n;i++)
            {
                if(a[j][i]>0)
                {
                    degree[i]--;
                    a[j][i]=-1;
                }
            }
        }
        if(jieshu==1)
        printf("YES\n");
        else
        printf("NO\n");
    }
}