POJ2485 Highways【Prim】

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.

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

Sample Output

Hint

Source

POJ Contest,Author:Mathematica@ZSU

题目大意：岛上要建高铁，有N个站点，给你一个图，表示这N个站点每个站点之间的距离，

要求建造的高铁路线能连接所有的站点，并且使总的路程最短。求出满足情况的路线中两个站

点间最长的路。

思路：根据要求求出最小生成树，并求出最小生成树上的最大边，就是最终答案。

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;

int Map[550][550],low[550],vis[550];

void Prim(int N)
{
    int MAX = 0;
    memset(vis,0,sizeof(vis));
    int pos = 1;
    int ans = 0;
    vis[1] = 1;
    for(int i = 1; i <= N; ++i)
        if(i != pos)
            low[i] = Map[pos][i];
    for(int i = 1; i < N; ++i)
    {
        int Min = 0xffffff0;
        for(int j = 1; j <= N; ++j)
        {
            if(!vis[j] && Min > low[j])
            {
                pos = j;
                Min = low[j];
            }
        }
        vis[pos] = 1;
        ans += Min;
        if(MAX < Min)
            MAX = Min;
        for(int j = 1; j <= N; ++j)
            if(!vis[j] && low[j] > Map[pos][j])
                low[j] = Map[pos][j];
    }
    printf("%d\n",MAX);
}
int main()
{
    int N,T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&N);
        for(int i = 1; i <= N; ++i)
            for(int j = 1; j <= N; ++j)
                scanf("%d",&Map[i][j]);
        Prim(N);
    }


    return 0;
}

POJ2485 Highways【Prim】

原文：http://blog.csdn.net/lianai911/article/details/43090125