poj_2485 Highways

最新推荐文章于 2021-08-01 08:48:49 发布

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最新推荐文章于 2021-08-01 08:48:49 发布

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本文介绍了一道关于寻找图中最小生成树最大边的经典算法题。通过Prim算法实现，确保所有节点相连的同时，最小化最长边的长度。

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Highways

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 16037	Accepted: 7460

题目链接：http://poj.org/problem?id=2485

Description

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

Flatopian towns are numbered from 1 to N. Each highway connects exactly twotowns. All highways follow straight lines. All highways can be used in bothdirections. Highways can freely cross each other, but a driver can only switchbetween 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 tobe built. However, they want to guarantee that every town is highway-reachablefrom every other town.

Input

The first line of input is aninteger T, which tells how many test cases followed.
The first line of each case is an integer N (3 <= N <= 500), which is thenumber 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 aninteger within [1, 65536]) between village i and village j. There is an emptyline after each test case.

Output

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

Sample Input

0 990 692

990 0 179

692 179 0

Sample Output

692

Hint

Huge input,scanf isrecommended.

Source

POJ Contest,Author:Mathematica@ZSU

题意：

给你一张图，让你求最小生成树中最大的边

解题思路：

直接用prim算法来求解即可。

代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#define MAX 503
#define VALUE 0xfffff
using namespace std;

int g[MAX][MAX];
char gra[MAX][7];
int minCost[MAX];
int visited[MAX];

int prim(int n)
{
    int i;
    for(i=0;i<n;i++)
    {
        visited[i]=0;
        minCost[i]=VALUE;
    }
    minCost[0]=0;
   // int res=0;
    int max=0;
    while(true)
    {
        int t=-1;
        for(i=0;i<n;i++)
        {
            if(visited[i]==0 && (t==-1 || minCost[i]<minCost[t]))
            {
                t=i;
            }
        }
        if(t==-1)
            break;
        visited[t]=1;
        if(minCost[t]>max)
        {
            max=minCost[t];
        }
       // res+=minCost[t];
        for(i=0;i<n;i++)
        {
            if(minCost[i]>g[i][t] && g[i][t]!=0)
            {
                minCost[i]=g[i][t];
            }
        }
    }
    return max;
}


int main()
{
    int ts;
    int i,j;
    scanf("%d",&ts);
    while(ts--)
    {
        memset(g,0,sizeof(g));
        int n;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                scanf("%d",&g[i][j]);
            }
        }
        printf("%d\n",prim(n));
    }
    return 0;
}