POJ 3687 拓扑排序应用

最新推荐文章于 2021-05-30 18:03:57 发布

原创最新推荐文章于 2021-05-30 18:03:57 发布 · 304 阅读

0 ·

CC 4.0 BY-SA版权

POJ 专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一个球排序问题的解决方法，通过使用优先队列实现拓扑排序算法来找到符合特定条件的球的排列顺序。该算法能够处理多达200个不同重量的球，并根据给定的约束条件确定每个球的标签。

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

1 2 3 4
-1
-1
2 1 3 4
1 3 2 4
#include <IOSTREAM>
#include <QUEUE>
#include <CSTDLIB>
using namespace std;

#define NUM 201

int gmap[NUM][NUM];
int N,M;
int degree[NUM];
int value[NUM];
priority_queue<int>  que;

bool tuopo()
{
	int remain=N;
	int ic;
	for (ic=1;ic<=N;ic++)
	{
		if (degree[ic]==0)
			que.push(ic);
	}
	if (que.size()==0)
		return false;
	while (que.size()!=0)
	{
		int node=que.top();
		value[node]=remain--;
		que.pop();
		for (ic=1;ic<=N;ic++)
		{
			if (gmap[ic][node]==1)
			{
				gmap[ic][node]=-1;
				degree[ic]--;
				if (degree[ic]==0)
					que.push(ic);
			}
		}
	}
	if (remain==0)
		return true;
	else
		return false;
}
int main()
{
	int a,b,T;
	cin>>T;
	while (T!=0)
	{
		memset(gmap,0,sizeof(gmap));
		memset(degree,0,sizeof(degree));
     	memset(value,0,sizeof(value));
		N=M=0;
		cin>>N>>M;
		while (M!=0)
		{
			cin>>a>>b;
			if(gmap[a][b]==0)
			{
				gmap[a][b]=1;
				degree[a]++;
			}
			M--;
		}
		if (tuopo())
		{
			for (int ic=1;ic<=N;ic++)
			{
				cout<<value[ic]<<" ";
			}
			cout<<endl;
		}
		else
			cout<<"-1"<<endl;		
		T--;
	}
	return 0;
}