poj 2485 Highways 最小生成树

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Highways

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19004		Accepted: 8815

Description

The island nation of Flatopia is perfectly flat. Unfortunately, Flatopia has no public highways. So the traffic is difficult in Flatopia. The Flatopian government is aware of this problem. They're planning to build some highways so that it will be possible
to drive between any pair of towns without leaving the highway system. 



Flatopian towns are numbered from 1 to N. Each highway connects exactly two towns. All highways follow straight lines. All highways can be used in both directions. Highways can freely cross each other, but a driver can only switch between highways at a town
that is located at the end of both highways. 



The Flatopian government wants to minimize the length of the longest highway to be built. However, they want to guarantee that every town is highway-reachable from every other town.

Input

The first line of input is an integer T, which tells how many test cases followed. 

The first line of each case is an integer N (3 <= N <= 500), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 65536]) between
village i and village j. There is an empty line after each test case.

Output

For each test case, you should output a line contains an integer, which is the length of the longest road to be built such that all the villages are connected, and this value is minimum.

Sample Input

1

3
0 990 692
990 0 179
692 179 0

Sample Output

692

Hint

Huge input,scanf is recommended.

题目大意是要在多个村庄之间修公路，给出公路之间的距离，求总花费最小的方案中费用最高的那条公路，很裸的最小生成树问题

这题我用了prim + priority_queue实现的

#include<stdio.h>
#include<string.h>
#include<queue>
#include<utility>
using namespace std;
int map[501][501];
int n;

int prime()
{
	bool flag[501] = {0};
	priority_queue<	pair<int, int> , vector<pair<int, int > > , greater<pair<int ,int > > > q;
	pair<int, int > p, new_p;
	p.first = 0;
	p.second = 1;
	q.push(p);
	int max = 0;
	int t = 1;
	int j;
	for(j = 1; j <= n ;j++)
	{
		p = q.top();
		q.pop();
		if(flag[p.second] == 1)
		{
			j--;
			continue;
		}
		if(max < p.first)
			max = p.first;
		flag[p.second] = 1;
		int i;
		for(i = 1 ; i <= n; i++)
		{
			if(map[p.second][i] != 0 && flag[i] == 0)
			{
				new_p.first = map[p.second][i];
				new_p.second = i;
				q.push(new_p);
			}
		}
	}
	return max;
}
int main()
{
	int t;
	scanf("%d", &t);
	while(t--)
	{
		memset(map, 0, sizeof(map));
		scanf("%d", &n);
		int i, j;
		for(i = 1; i <= n; i++)
		{
			for(j = 1; j <= n; j++)
			{
				scanf("%d", &map[i][j]);
			}
		}
		int f = prime();
		printf("%d\n", f);
	}

	return 0;
}