hdu-1162 Eddy's picture---浮点数的MST

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1162

题目大意：

给n个点，求MST权值

解题思路：

直接prim算法

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 100 + 10;
const int INF = 1e9 + 7;
double Map[maxn][maxn];
double lowcost[maxn];
int mst[maxn];
int n, m;
double prim(int u)
{
    double ans = 0;
    for(int i = 1; i <= n; i++)
    {
        lowcost[i] = Map[u][i];
        mst[i] = u;
    }
    mst[u] = -1;
    for(int i = 1; i < n; i++)
    {
        double minn = INF * 1.0;
        int v = -1;
        //寻找lowcost数组里面的未加入mst的最小值
        for(int j = 1; j <= n; j++)
        {
            if(mst[j] != -1 && lowcost[j] < minn)
            {
                v = j;
                minn = lowcost[j];
            }
        }
        if(v != -1)
        {
            mst[v] = -1;
            ans += lowcost[v];
            for(int j = 1; j <= n; j++)
            {
                if(mst[j] != -1 && lowcost[j] > Map[v][j])
                {
                    lowcost[j] = Map[v][j];
                    mst[j] = v;
                }
            }
        }
    }
    printf("%.2f
", ans);
}
struct node
{
    double x, y;
}a[maxn];
double dis(node a, node b)
{
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
int main()
{
    while(scanf("%d", &n) != EOF)
    {
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)Map[i][j] = INF * 1.0;
        memset(lowcost, 0, sizeof(lowcost));
        for(int i = 1; i <= n; i++)
        {
            scanf("%lf%lf", &a[i].x, &a[i].y);
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = i + 1; j <= n; j++)
            {
                double d = dis(a[i], a[j]);
                Map[i][j] = Map[j][i] = d;
            }
        }
        prim(1);

    }
    return 0;
}