poj3687——Labeling Balls（拓扑排序）

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

编号为a的球轻于b的球，求可能的重量从小到大的排序请况。越在前面的球编号尽量越小。
将轻重有关系的球建图，从编号由大到小遍历，如果有没有比它更重的球，那这个球就是最重的，并从图里去除相关的关系。

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <map>#include <cmath>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 205#define Mod 10001using namespace std;int n,m;int mp[MAXN][MAXN],indegree[MAXN],ans[MAXN];int main(){    int t,a,b,j,i;    scanf("%d",&t);    while(t--)    {        memset(mp,0,sizeof(mp));        memset(indegree,0,sizeof(indegree));        scanf("%d%d",&n,&m);        while(m--)        {            scanf("%d%d",&a,&b);            if(!mp[b][a])            {                mp[b][a]=1;                indegree[a]++;            }        }        for(i=n; i>=1; --i)        {            for(j=n; j>=1; --j)            {                if(indegree[j]==0)                {                    indegree[j]--;                    ans[j]=i;                    for(int k=1; k<=n; ++k)                        if(mp[j][k])                            indegree[k]--;                    break;                }            }            if(j<1)                break;        }        if(i>=1)            cout<<"-1"<<endl;        else        {            cout<<ans[1];            for(int i=2; i<=n; ++i)                cout<<" "<<ans[i];            cout<<endl;        }    }    return 0;}