拓扑排序之反向建树

Description

Windy has N balls of distinct weights from 1 unit to N units. Now he tries to label them with 1 to N in such a way that:

No two balls share the same label.
The labeling satisfies several constrains like "The ball labeled with a is lighter than the one labeled with b".
Can you help windy to find a solution?

Input

The first line of input is the number of test case. The first line of each test case contains two integers, N (1 ≤ N ≤ 200) and M (0 ≤ M ≤ 40,000). The next M line each contain two integers a and b indicating the ball labeled with a must be lighter than the one labeled with b. (1 ≤ a, b ≤ N) There is a blank line before each test case.

Output

For each test case output on a single line the balls' weights from label 1 to label N. If several solutions exist, you should output the one with the smallest weight for label 1, then with the smallest weight for label 2, then with the smallest weight for label 3 and so on... If no solution exists, output -1 instead.

Sample Input

5

4 0

4 1
1 1

4 2
1 2
2 1

4 1
2 1

4 1
3 2
Sample Output

1 2 3 4
-1
-1
2 1 3 4
1 3 2 4

 

题目大意：

     这道题每次输入a，b的时候表示的是编号为a的球比编号为b的球轻，最后输出的是从编号 1

     到编号 n每一个小球的重量，若是存在多组解，输出使最小重量尽可能排在前边的那组解，亦即 全部解中 1到 n号球的重量的字典序最小。。。。

 

#include<stdio.h>#include<string.h>#define N 1010int map[N][N];int indegree[N];int vis[N];int a[N];int main(){    int i,j,m,n,x,y,t,p,flag;    scanf("%d",&t);    while(t--)    {        flag=0;        scanf("%d%d",&n,&m);        memset(map,0,sizeof(map));        memset(indegree,0,sizeof(indegree));        memset(a,0,sizeof(a));        for(i=1; i<=m; i++)        {            scanf("%d%d",&x,&y);            if(map[x][y]==0)            {                map[x][y]=1;                indegree[x]++;//计算出度            }        }        p=n;        for(;;)//反向建树        {            for(i=n; i>=1; i--)                if(a[i]==0&&indegree[i]==0)                    break;            if(i==0)                break;            a[i]=p--;            for(j=1; j<=n; j++)                if(map[j][i]==1)                    indegree[j]--;        }        if(p==0)        {            for(i=1; i<n; i++)                printf("%d ",a[i]);            printf("%d\n",a[i]);        }        else            printf("-1\n");    }    return 0;}