POJ_2349_Arctic Network(最小生成树)

Arctic Network

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11782		Accepted: 3867

Description
The Department of National Defence (DND) wishes to connect several northern outposts by a wireless network. Two different communication technologies are to be used in establishing the network: every outpost will have a radio transceiver
and some outposts will in addition have a satellite channel.

Any two outposts with a satellite channel can communicate via the satellite, regardless of their location. Otherwise, two outposts can communicate by radio only if the distance between them does not exceed D, which depends of the power of the transceivers.
Higher power yields higher D but costs more. Due to purchasing and maintenance considerations, the transceivers at the outposts must be identical; that is, the value of D is the same for every pair of outposts.

Your job is to determine the minimum D required for the transceivers. There must be at least one communication path (direct or indirect) between every pair of outposts.
Input
The first line of input contains N, the number of test cases. The first line of each test case contains 1 <= S <= 100, the number of satellite channels, and S < P <= 500, the number of outposts. P lines follow, giving the (x,y)
coordinates of each outpost in km (coordinates are integers between 0 and 10,000).
Output
For each case, output should consist of a single line giving the minimum D required to connect the network. Output should be specified to 2 decimal points.
Sample Input

1
2 4
0 100
0 300
0 600
150 750

Sample Output
212.13

题意：这题是上acm课的时候老师布置的一道题，题意反正我是看很久没看懂。大概就是求最小生成树的第k大边，于是，就是裸题了。
分析：kruskal算法求最小生成树。第k大边，也就是第(n-k)小边，所以跑kruskal的时候到(n-k)条边时，直接输出边的权值即可。对了，该题在POJ上交的时候g++一直wa，然后改成c++才ac的。关于这点，我也是不清楚为什么，若有大神经过，望指点迷津。
题目链接：http://poj.org/problem?id=2349
代码清单：

#include<map>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<cctype>
#include<string>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;

const int maxn = 500 + 5;

struct Edge{
    int u;
    int v;
    double dis;
}edge[maxn*maxn];

struct Point{
    double x;
    double y;
}point[maxn];

int father[maxn];
int t,s,p,cnt,sum;

void init(){
    cnt=0;
    sum=0;
    for(int i=0;i<maxn;i++) father[i]=i;
}

void input(){
    scanf("%d%d",&s,&p);
    for(int i=0;i<p;i++)
        scanf("%lf%lf",&point[i].x,&point[i].y);
}

double Dis(Point p1,Point p2){  //求两点距离
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}

bool cmp(Edge a,Edge b){ return a.dis<b.dis; } //边按从小到大排序

int Find(int x){             //并查集压缩路径
    if(x!=father[x]) return father[x]=Find(father[x]);
    return father[x];
}

void solve(){
    for(int i=0;i<p-1;i++){
        for(int j=i+1;j<p;j++){
            edge[cnt].u=i;
            edge[cnt].v=j;
            edge[cnt++].dis=Dis(point[i],point[j]);
        }
    }
    sort(edge,edge+cnt,cmp);
    for(int i=0;i<cnt;i++){
        int uu=Find(edge[i].u);
        int vv=Find(edge[i].v);
        if(uu!=vv){
            sum++;
            father[vv]=uu;
            if(sum+s==p){
                printf("%.2lf\n",edge[i].dis);
                break;
            }
        }
    }
}

int main(){
    scanf("%d",&t);
    while(t--){
        init();
        input();
        solve();
    }return 0;
}