3687-Labeling Balls-反向拓扑排序

本文介绍了一个球排序问题的解决方法，通过输入不同重量球之间的比较关系，利用图论中的拓扑排序思想，实现了一种算法来确定球的正确排序方式。如果存在多个解，则输出按重量最小顺序排列的方案。

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

Sample Output

题意：假设原来球一样重，后面告诉的条件，例如2 1 前面的球比后面的球重，把球按照重量排序，重的在后面且序号较大，然后输出每个球的位置，没有处理的球在原来的位子，反向排，从n到1

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int n,m;
int ls[500];
int du[250];
int map[250][250];
void tuo()
{
   bool flag ;
   int c = 0;
   int mm;
   for(int i = n; i >= 1; i--)
   {
       flag = false;
       for(int j = n; j >= 1; j--)
       {
           if(du[j]==0)
           {
               mm = j;
               flag = true;
               break;
           }
       }
       if(!flag)
       {
           printf("-1\n");
          return;
       }
       ls[mm] = i;
       du[mm]=-1;
       for(int j = 1;j <= n; j++)
       {
           if(map[mm][j])
            du[j]--;
       }
   }
   for(int i = 1; i <= n; i++)
   {
       printf("%d",ls[i]);
       if(i < n)
        printf(" ");
   }
   printf("\n");

}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
       scanf("%d%d",&n,&m);
       memset(du,0,sizeof(du));
       memset(map,0,sizeof(map));
       for(int i = 0; i < m; i++)
       {
           int u,v;
           scanf("%d%d",&u,&v);
           if(!map[v][u])
           {
               map[v][u]=1;
               du[u]++;
           }
       }
       tuo();
    }
    return 0;
}