poj 3625 Building Roads

Building Roads
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 11902 Accepted: 3395
Description

Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms.

Each of the N (1 ≤ N ≤ 1,000) farms (conveniently numbered 1..N) is represented by a position (Xi, Yi) on the plane (0 ≤ Xi ≤ 1,000,000; 0 ≤ Yi ≤ 1,000,000). Given the preexisting M roads (1 ≤ M ≤ 1,000) as pairs of connected farms, help Farmer John determine the smallest length of additional roads he must build to connect all his farms.

Input

• Line 1: Two space-separated integers: N and M
• Lines 2..N+1: Two space-separated integers: Xi and Yi
• Lines N+2..N+M+2: Two space-separated integers: i and j, indicating that there is already a road connecting the farm i and farm j.

Output

• Line 1: Smallest length of additional roads required to connect all farms, printed without rounding to two decimal places. Be sure to calculate distances as 64-bit floating point numbers.

Sample Input

4 1
1 1
3 1
2 3
4 3
1 4
Sample Output

4.00

最小生成树 事先已经有连在一起的农场了,只需要把这些农场的边设成0，跑一遍prime就好了

#include <stdio.h>#include <algorithm>#include <math.h>#include <string.h>#define maxn 1100#define INF 999999999double edge[maxn][maxn];bool vis[maxn];double lowcost[maxn];using namespace std;int n,m;struct node{    double x,y;}a[maxn];double dis(node a,node b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}double prime(int num){    vis[num] = 1;    double sum = 0;    for(int i = 1; i <= n; i++)    {        lowcost[i] = edge[i][num];    }    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;        sum += minn;        for(int j = 1; j <= n; j++)        {            if(vis[j] == 0 && lowcost[j] > edge[j][temp])            {                lowcost[j] = edge[j][temp];            }        }    }    return sum;}int main(){    while(~scanf("%d%d",&n,&m))    {        memset(vis,0,sizeof(vis));        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] = dis(a[i],a[j]);            }        }        for(int i = 1; i <= m; i++)        {            int u,v;            scanf("%d%d",&u,&v);            edge[u][v] = edge[v][u] = 0.0;        }        double ans = prime(1);        printf("%.2f\n",ans);    }}