UVA Ordering Tasks

John has n tasks to do. Unfortunately, the tasks are not independent and the execution of one task is
only possible if other tasks have already been executed.
Input
The input will consist of several instances of the problem. Each instance begins with a line containing
two integers, 1 ≤ n ≤ 100 and m. n is the number of tasks (numbered from 1 to n) and m is the
number of direct precedence relations between tasks. After this, there will be m lines with two integers
i and j, representing the fact that task i must be executed before task j.
An instance with n = m = 0 will finish the input.
Output
For each instance, print a line with n integers representing the tasks in a possible order of execution.
Sample Input
5 4
1 2
2 3
1 3
1 5
0 0
Sample Output
1 4 2 5 3

题目是说John有n个任务要做，但是这些任务并不是相互独立的，而是要先做完某个任务才能做下一个任务

拓扑排序模板题。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int a[510][510];
int num[510];
int book[510];
int main(){
	int n,m;
	while(~scanf("%d%d",&n,&m)){
		if(!n&&!m) break;
		memset(a,0,sizeof(a));
		memset(num,0,sizeof(num));
		memset(book,0,sizeof(book));
		while(m--){
			int x,y;
			cin>>x>>y;
			//两点之间的边加一 
			a[x][y]++;
			//入度加一 
			num[y]++;
		}
		for(int i=1;i<=n;i++){
			int j=1;
			//寻找下一个入度不为零的点 
			while(num[j]!=0) j++;
			//将其变为-1 
			num[j]=-1;
			//记录这个点 
			book[i]=j;
			for(int k=1;k<=n;k++){
				//删除此点之后要将与他相关联的边删掉
				//与他相邻的点入度-1 
				if(a[j][k]){
					num[k]--;
				}
			}
		}
		for(int i=1;i<=n-1;i++) cout<<book[i]<<' ';
		cout<<book[n]<<endl;
	}
}