1002 ProblemB

题意：
给出点的坐标求连通所有的点画的最短线，结果保留两位小数。
思路：
知道坐标就可以求出两点间距离，然后用最小生成树做。

// problemB.cpp : 定义控制台应用程序的入口点。
//

//#include "stdafx.h"
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iomanip>
using namespace std;
double path[102][102];
int flag[102];
double closedge[102];
double ma;
double cnt;

typedef struct {
    double x, y;
}input;

input temp[102];

double calculate(double x1, double y1, double x2, double y2)
{
    return sqrt((x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2));
}

double CreatMST(int n)
{
    int i, j, x;
    double k;
    flag[0] = 1;
    for (i = 1; i<n; ++i)
        closedge[i] = path[0][i];
    for (i = 1; i<n; ++i)
    {
        k = ma, x = 1;
        for (j = 1; j<n; ++j)
            if (!flag[j] && closedge[j] < k)
                x = j, k = closedge[j];
        flag[x] = 1;
        cnt += k;
        for (j = 1; j<n; ++j)
            if (!flag[j] && closedge[j] > path[x][j])
                closedge[j] = path[x][j];
    }
    return cnt;
}

int main()
{
    int i, j, k, t, x, y, n, m;
    while (cin>>n)
    {
        ma = 0;
        cnt = 0;
        memset(temp, 0, sizeof(input) * 102);
        memset(flag, 0, sizeof(flag));
        memset(closedge, 0, sizeof(closedge));
        memset(path, 0, sizeof(double) * 102 * 102);
        for (i = 0; i<n; ++i)
            cin>>temp[i].x>>temp[i].y;

        for (i = 0; i<n; ++i)
            for (j = 0; j<n; ++j)
            {
                path[i][j] = calculate(temp[i].x, temp[i].y, temp[j].x, temp[j].y);
                if (ma < path[i][j]) ma = path[i][j];
            }
        cout << setiosflags(ios::fixed) << setprecision(2) << CreatMST(n) << endl;
    }
    return 0;
}