POJ 3687：Labeling Balls（优先队列+拓扑排序）

id=3687">Labeling Balls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10178		Accepted: 2815

Description

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

1. No two balls share the same label.
2. The labeling satisfies several constrains like "The ball labeled witha is lighter than the one labeled withb".

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 nextM line each contain two integersa and
b indicating the ball labeled witha must be lighter than the one labeled withb. (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 labelN. 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

在主要的拓扑排序的基础上又添加了一个要求：编号最小的节点要尽量排在前面；在满足上一个条件的基础上，编号第二小的节点要尽量排在前面；

在满足前两个条件的基础上。编号第三小的节点要尽量排在前面……依此类推。点击打开链接又是看结题报告。。。

。哎。。

。。

太弱了

第一百篇。。留念。。。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>

using namespace std;

const int M = 250 ;
int t, n, m;
int outint[M];
int out[M];
int in[M];
int cut;
vector<int>amap[M];
int flag;

bool toposort()
{
    cut = 0;
    priority_queue<int>que;
    for(int i=1; i<=n; i++)
        if( !in[i] )
        que.push(i);
    while( !que.empty() )
    {
        int u = que.top();
        que.pop();
        outint[ cut++] = u;
        for( int i=0; i<amap[u].size(); i++ )
        {
            int v = amap[u][i];
            if( --in[v]==0 )
                que.push(v);
        }
    }
    if( cut<n )
        return false;
    else
        return true;
}

int main()
{
    scanf( "%d", &t );
    while( t-- )
    {
        memset( in, 0, sizeof(in) );
        scanf( "%d%d", &n, &m );
        for( int i=1; i<=n; i++ )
            amap[i].clear();
        for( int i=1; i<=m; i++ )
        {
            int a, b;
            scanf( "%d%d", &a, &b );
            amap[b].push_back(a);
            in[a]++;
        }
        if( !toposort() )
            printf("-1\n");
        else
        {
            for( int i=0; i<n; i++ )
                out[ outint[i] ] = n-i;
            for(int i=1; i<=n; i++)
            {
                if(i<n)
                    printf( "%d ", out[i] );
                else
                    printf( "%d\n", out[i] );
            }
        }
    }

    return 0;
}