uva 10305 Ordering Tasks

原题：
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
大意:
给你一堆有依赖的任务，让你排序。

#include <bits/stdc++.h>using namespace std;int n,m;int deg[101];int gra[101][101];vector<int> ans;void topsort(){    ans.clear();    int k;    for(int i=1;i<=n;i++)    {        for(int j=1;j<=n;j++)        {            if(deg[j]==0)            {                ans.push_back(j);                deg[j]--;                k=j;                break;            }        }        for(int j=1;j<=n;j++)        {            if(gra[k][j])            {                gra[k][j]=0;                deg[j]--;            }        }    }}int main(){    ios::sync_with_stdio(false);    while(cin>>n>>m,n+m)    {        memset(gra,0,sizeof(gra));        memset(deg,0,sizeof(deg));        for(int i=0;i<m;i++)        {            int a,b;            cin>>a>>b;            if(gra[a][b]==0)            {                gra[a][b]=1;                deg[b]++;            }        }        topsort();        int len=ans.size();        for(int i=0;i<len;i++)        {            if(i!=len-1)                cout<<ans[i]<<" ";            else                cout<<ans[i]<<endl;        }    }    return 0;}

解答：
裸的拓扑排序问题~ 直接模板就能交过