POJ 3687 Labeling Balls (拓扑排序)

Labeling Balls
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu
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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 Mline 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

先将编号经过拓扑排序找出，然后再赋予他们重量，最后按序号大小输出重量，注意防止重边现象

#include <cstdio>  
#include <cstring>  
#include <queue>  
#define maxn 100002  
using namespace std;  
int head[maxn], indegree[maxn], ans[maxn],j;  
struct Node{  
    int to, next;  
} map[maxn];  
  
void topoSort(int n)  
{   j=n;
    priority_queue<int> Q;  
    int i, u;  
    for(i = 1; i <= n; ++i)  
        if(!indegree[i])
Q.push(i);  
    while(!Q.empty())
{  
        u=Q.top();  
         Q.pop();
ans[u]=j;
j--;  
        for(i = head[u]; i != -1; i = map[i].next)  
            if(!--indegree[map[i].to])
Q.push(map[i].to);          
    }      
   
}  
  
int main()  
{  
    int t, n, m, a, b, i;  
    scanf("%d", &t);  
    while(t--){  
        memset(indegree, 0, sizeof(indegree));  
        memset(head, -1, sizeof(head));  
        scanf("%d%d", &n, &m);  
        for(i = 0; i < m; ++i){  
            scanf("%d%d", &a, &b);  
            map[i].to = a;  
            map[i].next = head[b];  
            head[b] = i;  
            ++indegree[a];  
        }  
        topoSort(n); 
if(j==0)
{
   for(i = 1; i <= n; i++)   
        if(i != n)
printf("%d ", ans[i]);  
        else printf("%d\n", ans[i]);  
} 
else
printf("-1\n");
    }  
    return 0;  
}