poj 3625 Building Roads-Prim(最小生成树)

                                                                                                            Building Roads

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7756 Accepted: 2232

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 (X_i,Y_i) on the plane (0 ≤X_i ≤ 1,000,000; 0 ≤Y_i ≤ 1,000,000). Given the preexistingM 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: X_i andY_i
* Lines N+2..N+M+2: Two space-separated integers: i and j, indicating that there is already a road connecting the farmi and farmj.

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 11 13 12 34 31 4

Sample Output

4.00

 

好吧，。这题比较强大，不过我还是用不变的Prim模版做的。。哈哈。

先输入n,m。下面的n行，是输入n行坐标（x,y），在就是输入m行的（i,j）//

代码：

#include<cstdio>#include<cmath>#include<string>#include<cstring>#include<algorithm>#include<iostream>#include<cstring>using namespace std;double road[1010][1010];double dist[1010];int vis[1010];int n,m,i,j,rec;double d,sum,max;struct node{double x,y;};struct node data[1010];void init(){while(m--){scanf("%d %d",&i,&j);road[i][j]=road[j][i]=0;;}}void solve(){for(i=1;i<=n;i++){for(j=i+1;j<=n;j++){d=sqrt((data[i].x-data[j].x)*(data[i].x-data[j].x)+(data[i].y-data[j].y)*(data[i].y-data[j].y));road[i][j]=road[j][i]=d;}}}double Prim(){memset(vis,0,sizeof(vis));double sum=0.0; int k;for(int i=1;i<=n;i++)dist[i]=road[1][i];vis[1]=1;                for(int i=1;i<n;i++){double min=100000000.0;for(int j=1;j<=n;j++)if(!vis[j]&&dist[j]<min){min=dist[j];k=j;}sum+=min;vis[k]=1;for(int j=1;j<=n;j++){if(!vis[j]&&road[k][j]<dist[j])dist[j]=road[k][j];}}return sum;}int main(){while(scanf("%d %d",&n,&m)!=EOF){for(i=1;i<=n;i++)scanf("%lf %lf",&data[i].x,&data[i].y);memset(road,0,sizeof(road));solve();init();printf("%.2lf\n",Prim());}return 0;}