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poj 2485 Highways（最小生成树）

2024-07-18 06:31:33   230人阅读

题目链接：http://poj.org/problem?id=2485


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.

Source

POJ Contest,Author:Mathematica@ZSU


题意：

寻找最小生成树图中最长的路径！

和POJ 1258一样的，加个判断条件就可以了！

但是注意这题用memset 初始化visited会WA， 用for循环就不会了！POJ 1258 则不存在这个问题！


代码如下：

#include <cstdio>
#include <cstring>
#define INF 0x3f3f3f3f
#define MAXN 517
//创建m二维数组储存图表，low数组记录每2个点间最小权值，visited数组标记某点是否已访问
int m[MAXN][MAXN], low[MAXN], visited[MAXN];
int n;
int prim( )
{
    int i, j;
    int maxx = 0;
    int pos, minn, result=0;
    // memset(visited,0,sizeof(visited));
    for(i = 1; i <= n; i++)
        visited[i] = 0;
    visited[1] = 1;
    pos = 1;          //从某点开始，分别标记和记录该点
    for(i = 1; i <= n; i++)     //第一次给low数组赋值
    {
        if(i != pos)
            low[i] = m[pos][i];
        else
            low[i] = 0;
    }
    for(i = 1; i < n; i++) //再运行n-1次
    {
        minn = INF;   //找出最小权值并记录位置
        for(j = 1; j <= n; j++)
        {
            if(visited[j]==0 && minn>low[j])
            {
                minn = low[j];
                pos = j;
            }
        }
        // result += minn;   //最小权值累加
        if(minn > maxx)
            maxx = minn;
        visited[pos] = 1;   //标记该点
        for(j = 1; j <= n; j++)   //更新权值
            if(visited[j]==0 && low[j]>m[pos][j])
                low[j] = m[pos][j];
    }
    //return result;
    return maxx;
}
int main()
{
    int i,j,ans;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        memset(m,INF,sizeof(m));   //所有权值初始化为最大
        for(i = 1; i <= n; i++)
        {
            for(j = 1; j <= n; j++)
            {
                scanf("%d",&m[i][j]);
            }
        }
        ans = prim( );
        printf("%d\n",ans);
    }
    return 0;
}



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