文章标题 POJ 2485 ： Highways（最小生成树--kruskal+并查集）

Highways

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.
题意：要给n个城镇建铁路，使得花费最小，用一个矩阵给出每两个城镇建一条铁路需要的花费。要我们求出将这些城镇连起来中最小花费，将这写铁路中最长的一条输出来
分析：花费通过矩阵已经给我们了，所以只需要将这些路排一下序，直接用kruskal得到最小生成树，最后一条边就是所求的最长的铁路
代码：

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<vector>#include<math.h>#include<map>#include<queue> #include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;int n;int mp[505][505];int fa[505];//并查集int find(int x){    return fa[x]==x?x:fa[x]=find(fa[x]);}struct node {    int x,y;//铁路连的两个城镇    int dis;//距离};node a[250005];bool cmp(node a,node b){    return a.dis<b.dis;}int main (){    int t;    scanf ("%d",&t);    while (t--){        scanf ("%d",&n);        for (int i=1;i<=n;i++){            fa[i]=i;            for (int j=1;j<=n;j++){                scanf ("%d",&mp[i][j]);            }        }        int cnt=0;        for (int i=1;i<=n-1;i++){            for (int j=i+1;j<=n;j++){//找出所有城镇亮亮相连的边保存起来                a[cnt].x=i;                a[cnt].y=j;                a[cnt++].dis=mp[i][j];            }        }        sort (a,a+cnt,cmp);//排序        int connect=0;        int ans=0;        for(int i=0;i<cnt;i++){            int x=a[i].x;            int y=a[i].y;            int fx=find(x),fy=find(y);            if (fx!=fy){                connect++;                fa[fx]=fy;                ans=max(ans,a[i].dis);            }            if (connect==n-1){// 单找到n-1条边时结束                break;            }        }        printf ("%d\n",ans);    }    return 0;}