Labeling Balls 分类： POJ ...

最新推荐文章于 2024-12-16 13:21:50 发布

转载最新推荐文章于 2024-12-16 13:21:50 发布 · 115 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/juechen/p/4721957.html

本文解析了一道关于球排序的问题，通过拓扑排序算法解决球的权重与标签之间的约束关系，实现合理的球序排列。文章提供了完整的代码示例，并解释了如何处理多种可能的情况。

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

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("\n");
            }
            else
            {
                printf("-1\n");
            }
        }

    }
    return 0;
}