poj 2349 Arctic Network

Arctic Network
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 23234 Accepted: 7139
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： 有m个卫星和n个哨岗，装卫星的村庄可以自由联络不受限制，其余n-m个村庄只能通过radio连接，为了方便维护每个radio的接收范围都是一致的，求每个radio的最小范围。

做法： n个哨岗存在n-1条边，用prime算法跑一下最小生成树，开一个数组存一下每条边的长度，然后用卫星替代最长的边，剩下的边中最长的即为所求。

这题写完之后怎么交都是WA,搞了半天最后才知道是输出的问题，在G++下提交double 格式是%f,下面附一张区别的图

#include <algorithm>#include <stdio.h>#include <string.h>#include <cmath>using namespace std;const int maxn = 550;double edge[maxn][maxn];double lowcost[maxn];bool vis[maxn];double length[maxn];double INF = 1000000000;int n,m;struct node{    double x,y;}a[maxn];void prim(int num){    int p = 0;    vis[num] = 1;    for(int i = 1; i <= n; i++)    {        lowcost[i] = edge[num][i];    }    for(int i = 1; i < n; i++)    {        double minn = INF;        int temp;        for(int j =  1; j <= n; j++)        {            if(vis[j] == 0 && lowcost[j] < minn)            {                minn = lowcost[j];                temp = j;            }        }        vis[temp] = 1;        length[p++] = minn;        for(int j = 1; j <= n; j++)        {            if(lowcost[j] > edge[temp][j])            {                lowcost[j] = edge[temp][j];            }        }    }    sort(length,length+p);    printf("%.2f\n",length[p-m]);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        memset(vis,0,sizeof(vis));        scanf("%d%d",&m,&n);        for(int i = 1; i <= n; i++)        {            scanf("%lf%lf",&a[i].x,&a[i].y);        }        for(int i = 1; i <= n; i++)        {            for(int j = 1; j <= n; j++)            {                edge[i][j] = sqrt((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y));            }        }        prim(1);    }    return 0;}