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  • Labeling Balls 分类: POJ 2015-07-28 19:47 10人阅读 评论(0) 收藏

    Labeling Balls
    Time Limit: 1000MS Memory Limit: 65536K
    Total Submissions: 11893 Accepted: 3408

    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
    有一点不太明白,为什么从小到大就WA

    #include <iostream>
    #include <cmath>
    #include <cstring>
    #include <cstdlib>
    #include <stdio.h>
    #include <string>
    #include <stack>
    #include <queue>
    #include <algorithm>
    #include <map>
    #define WW freopen("a1.txt","w",stdout)
    
    using namespace std;
    
    const int INF = 0x3f3f3f3f;
    
    int n,m;
    
    int Du[250];
    
    int a[250];
    
    bool Map[250][250];
    
    bool vis[250];
    int  Topo()拓扑排序
    {
        int wight=n;
    
        memset(vis,false,sizeof(vis));
        for(int i=1; i<=n; i++)
        {
            int ans=0;
            for(int j=n; j>0; j--)
            {
                if(!vis[j]&&Du[j]==0)
                {
                    ans=j;
                    break;
                }
            }
            if(!ans)
            {
                break;
            }
            vis[ans]=true;
            a[ans]=wight--;
            for(int j=1; j<=n; j++)
            {
                if(!vis[j]&&Map[j][ans])
                {
                    Du[j]--;
                }
            }
        }
        return wight;
    }
    
    int main()
    {
        int T;
        int u,v;
        bool flag;
        scanf("%d",&T);
    
        while(T--)
        {
            scanf("%d %d",&n,&m);
            flag=true;
            memset(Map,false,sizeof(Map));
            memset(Du,0,sizeof(Du));
            for(int i=1; i<=m; i++)
            {
                scanf("%d %d",&u,&v);
                if(!Map[u][v])
                {
                    Map[u][v]=true;
                    Du[u]++;
                }
            }
            if(flag)
            {
                if(Topo()==0)
                {
                    for(int i=1; i<=n; i++)
                    {
                        if(i!=1)
                        {
                            printf(" ");
                        }
                        printf("%d",a[i]);
                    }
                    printf("
    ");
                }
                else
                {
                    printf("-1
    ");
                }
            }
    
        }
        return 0;
    }
    

    版权声明:本文为博主原创文章,未经博主允许不得转载。

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  • 原文地址:https://www.cnblogs.com/juechen/p/4721957.html
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