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poj_1125 Floyd算法简单应用

2014-07-25 12:52 309 查看
题目很简单,floyd很容易记忆,理解所谓的“松弛”就行,变量依次为k,i,j。

求图G中任意两点的最短路径;对每一个stockbroker看是否能传播给其他人信息,若都能,找到min_time,最后比较若干个stockbroker的min_time,找到最小的那个。

#include<iostream>
#include<fstream>

using namespace std;

const int MAX = 0xfffffff;
int dist[101][101];//max map

int n;//number of stockbrokers

void floyd()
{
	for (int k = 1; k <= n; k++)
	{
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				if (dist[i][j] > dist[i][k] + dist[k][j])
				{
					dist[i][j] = dist[i][k] + dist[k][j];
				}
			}
		}
	}
}

int main()
{
	//ifstream in("text.txt");
	while (cin >> n && n!=0)
	{
		//init
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				if (i != j)
					dist[i][j] = MAX;
				else
					dist[i][j] = 0;
			}
		}
		//input
		for (int i = 1; i <= n; i++)
		{
			int pair;
			cin >> pair;
			for (int j = 1; j <= pair; j++)
			{
				int client, time;
				cin >> client >> time;
				dist[i][client] = time;
			}
		}
		floyd();
		//handle start client & min time
		int ans=MAX,tmin;
		bool disjoint;
		int start;
		for (int i = 1; i <= n; i++)
		{
			tmin = -1;
			disjoint = false;
			for (int j = 1; j <= n && !disjoint; j++)
			{
				if (i!=j && dist[i][j]==MAX)
				{
					disjoint = true;
				}
				if (i != j && dist[i][j] > tmin)
				{
					tmin = dist[i][j];
				}
			}
			if (!disjoint && tmin<ans)
			{
				disjoint = false;
				ans = tmin;
				start = i;
			}
		}
		//output
		if (ans == MAX)
			cout << "disjoint" << endl;
		else
			cout << start << " " << ans<<endl;

	}
	//system("pause");
	return 0;
}
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