POJ 3625 Building Road（Prim）

Building Roads

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9435 Accepted: 2704

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 11 13 12 34 31 4

Sample Output

4.00

Source

USACO 2007 December Silver

#include <iostream>#include <cmath>#include <cstdio>using namespace std;#define MAXV 1005#define INF 1e12const int start = 1;bool visit[MAXV] = { false };double dist[MAXV];int preNode[MAXV];double graph[MAXV][MAXV];int vertex_size;int road_size;struct Point{    double x;    double y;};Point pointArr[MAXV];double calDist( Point a, Point b ){        return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));    }void prim(){        double res = 0.0;    int nextIndex;    int temp_node_num = vertex_size;        for( int index = 1; index <= vertex_size; ++index )        dist[index] = graph[1][index];    for( int count_num = 1; count_num <= vertex_size; ++count_num ){                double  minDistance = INF;        int i;                for( i = 1; i <= vertex_size; ++i ){            if( minDistance > dist[i] && !visit[i] ){                                minDistance = dist[i];                nextIndex = i;                            }        }                visit[nextIndex] = true;             for( i = 1; i <= vertex_size; ++i ){            if( graph[nextIndex][i] < dist[i] && !visit[i] )                dist[i] = graph[nextIndex][i];        }    }        for( int i = 1; i <= vertex_size; ++i ){        if( dist[i] != -1 )             res += dist[i];    }        printf( "%.2lf\n", res );    }void init(){        cin >> vertex_size >> road_size;    int i, j;        for( i = 1; i <= vertex_size; ++i )                cin>>pointArr[i].x>>pointArr[i].y;    for( i = 1; i <= vertex_size; ++i ){        for( j = 1; j <= vertex_size; ++j ){                        graph[i][j] = graph[j][i] = calDist( pointArr[i], pointArr[j] );                    }    }        for( i = 1; i <= road_size; ++i ){                int startPoint, endPoint;        cin >> startPoint >> endPoint;        graph[startPoint][endPoint] = -1;        graph[endPoint][startPoint] = -1;            }}int main(){        init();    prim();        return 0;    }