poj 3687 Labeling Balls

最新推荐文章于 2019-06-05 14:06:51 发布

转载最新推荐文章于 2019-06-05 14:06:51 发布 · 99 阅读

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原文链接：http://www.cnblogs.com/372465774y/archive/2012/11/12/2766302.html

文章讨论了一个关于球权重排列的问题，给出了求解方法并提供了代码实现。通过反向建图和优先队列来找到满足条件的权重分配。

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Labeling Balls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8225		Accepted: 2234

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

Sample Output

Source

POJ Founder Monthly Contest – 2008.08.31, windy7926778

//我开始的思路是有点对、只是贪心反了、这个输出的是重量呀、唉、、英语不好

//反向建图，然后优先选择编号大的给重量大的、、

#include <iostream>
#include <stdio.h>
#include <queue>
#include <stack>
#include <set>
#include <vector>
#include <math.h>
#include <string.h>
#include <algorithm>
using namespace std;
vector <int> v[202];
int in[202];
int pt[202];
int main()
{
    int T;
    int n,m;
    int a,b;
    int i,k;
    int q;
    scanf("%d",&T);
    while(T--)
    {
        priority_queue<int > Q;
        scanf("%d %d",&n,&m);
        memset(in,0,sizeof(in));
        q=n;
        for(i=0;i<m;i++)
        {
            scanf("%d %d",&a,&b);
            v[b].push_back(a);
            in[a]++;
        }
        for(i=1;i<=n;i++)
         if(in[i]==0)
          Q.push(i);
        int l;
        while(!Q.empty())
        {
            k=Q.top();Q.pop();
            pt[k]=q--;
            l=v[k].size();
            for(i=0;i<l;i++)
             if(--in[v[k][i]]==0)
              Q.push(v[k][i]);
        }
        for(i=1;i<=n;i++)
         if(in[i])
          break;
        if(i<=n) printf("-1\n");
        else
        {
             printf("%d",pt[1]);
            for(i=2;i<=n;i++)
             printf(" %d",pt[i]);
             printf("\n");
        }
        for(i=1;i<=n;i++) v[i].clear();
    }
    return 0;
}

转载于:https://www.cnblogs.com/372465774y/archive/2012/11/12/2766302.html