POJ 2485 Highways

最新推荐文章于 2017-03-28 13:58:39 发布

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本文介绍了一个经典的最小生成树问题，并通过克鲁斯卡尔算法实现了解决方案。该问题旨在找到连接所有节点的最短路径，确保任意两镇间都能通过高速公路系统相连，同时最小化最长高速公路的长度。

A - Highways

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status

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

Sample Output

Hint

Huge input,scanf is recommended.

最小生成树第一发。克鲁斯卡尔水过。手敲了一遍。

#include <stdio.h>
#include <algorithm>
#define N 505
using namespace std;
typedef struct
{
        int x,y,d;
}Node;
int fa[N];
Node graph[N*N];
int input[N][N];
bool cmp(Node a,Node b)
{
        return a.d<b.d;
}
void InitSet(int n)
{
    for(int i=1; i<=n; i++) fa[i] = i ;
}
int Find(int x)
{
    return fa[x] == x ? x : fa[x] = Find(fa[x]) ;
}
bool Merge(int u ,int v)
{
    int fu = Find(u) , fv = Find(v) ;
    if(fu != fv) fa[fv] = fu ;
    return fu != fv ;
}
int main()
{
        int t;
        scanf("%d",&t);
        while(t--)
        {
                int n,ans=0;
                scanf("%d",&n);
                InitSet(n);
                int k=1;
                for(int i=1;i<=n;i++)
                {
                        for(int j=1;j<=n;j++)
                                scanf("%d",&input[i][j]);
                        for(int j=1;j<i;j++)
                                graph[k].x=i,graph[k].y=j,graph[k++].d=input[i][j];
                }
                sort(graph,graph+k,cmp);
                for(int i=1;i<k;i++)
                        if(Merge(graph[i].x,graph[i].y))
                                ans=graph[i].d;
                printf("%d\n",ans);
        }
        return 0;
}