POJ3625 Building Roads

Building Roads Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11366 Accepted: 3224

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 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: X_i and Y_i  * 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

Source

USACO 2007 December Silver

思路：实际上这道题并不是很难，然后通过这道题目有学到了一个新的技能，用c++提交double输出%lf，G++提交double输出%f。

#include <cstdio> #include <cmath> #include <cstring> #define INF 0x3f3f3f3f using namespace std; struct node{ double x,y; }point[1005]; long long n,m; long long vis[1005]; double dis[1005],a[1005][1005],ans=0; double distance(double x1, double y1, double x2, double y2){ return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2)); } void Prim(){ for(int i = 1; i <= n; i ++){ dis[i] = a[1][i]; vis[i] = 0; } vis[1] = 1; double _min; int k; for(int i = 1; i < n; i ++){ _min = INF; k = 0; for(int j = 1; j <= n; j ++){ if(!vis[j] && dis[j] < _min){ _min = dis[j]; k = j; } } ans += _min; vis[k] = 1; for(int j = 1; j <= n;j ++){ if(!vis[j] && dis[j] > a[k][j]) dis[j] = a[k][j]; } } } int main(){ while(~scanf("%lld%lld",&n,&m)){ for(int i = 1; i <= n; i ++) for(int j = 1; j <= n; j ++) a[i][j] = i == j ? 0 : INF; int u,v; for(int i = 1; i <= n; i ++){ scanf("%lf%lf",&point[i].x,&point[i].y); } for(int i = 1; i <= n; i ++){ for(int j = i + 1; j <= n; j ++) a[i][j] = a[j][i] = distance(point[i].x,point[i].y,point[j].x,point[j].y); } for(int i = 1; i <= m; i ++){ scanf("%d%d",&u,&v); a[u][v] = a[v][u] = 0; } ans = 0; Prim(); printf("%.2lf\n",ans); } return 0; }