Building Roads(3625)

Building Roads

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5086		Accepted: 1485

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

呵呵，这个题目就是最小生成树的变种。只是我交了几次才AC。WA原因在代码后面有注释！

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
#define size 1008
#define MAX 99999999
double tree[size][size],x[size],y[size]; 
double low[size];
bool flag[size]={1};
double prim(int n)
{
    double sum = 0;
	int i,k,time=1;
    for(i = 1; i < n; i++)
        low[i] = tree[0][i];
    while(time<=n-1)
	{
        double min = MAX;
        int j = 0;
        for(k = 1; k < n; k++)
            if( low[k] < min && flag[k]==false )
            {
                min = low[k];
                j = k;
            }
        sum += min;
        flag[j] = true;
        for(k = 1; k < n; k++)
            if(tree[j][k] < low[k]&&flag[k]==false)
                low[k] = tree[j][k];
			time++;
    }
    return sum;
}
int main()
{       

	int n,m,i=0,j,num,c,d;
	double xx,yy,a,b;
	cin>>n>>m;
	num=n;
	while(n--)
	{
		cin>>a>>b;
		x[i]=a;
		y[i]=b;
		i++;
	}
	for(i=0;i<num;i++)
		for(j=0;j<num;j++)
		{
			if(i==j)//刚开始的时候这条语句我写成了赋值语句，寒！！！！！！
				tree[i][i]=MAX;//自己与自己不能联通，距离把他设为+∞
			else
			{
				xx=fabs(x[i]-x[j]);
				yy=fabs(y[i]-y[j]);
				tree[i][j]=sqrt( xx*xx+yy*yy );//各个村庄的距离
			}
		}
	
	
	while(m--)
	{
		cin>>c>>d;
		tree[c-1][d-1]=0;
		tree[d-1][c-1]=0;//标记已经建好的路，我就是这条语句没有加上去，所以WA了几次
	}
	n=num;
	cout<<setprecision(2)<<fixed<<prim(n)<<endl;
    return 0;
}