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 =  + ;
 const int INF = 1e9 + ;
 double Map[maxn][maxn];
 double lowcost[maxn];
 int mst[maxn];
 int n, m;
 double prim(int u)
 {
     double ans = ;
     for(int i = ; i <= n; i++)
     {
         lowcost[i] = Map[u][i];
         mst[i] = u;
     }
     mst[u] = -;
     for(int i = ; i < n; i++)
     {
         double minn = INF * 1.0;
         int v = -;
         //寻找lowcost数组里面的未加入mst的最小值
         for(int j = ; j <= n; j++)
         {
             if(mst[j] != - && lowcost[j] < minn)
             {
                 v = j;
                 minn = lowcost[j];
             }
         }
         if(v != -)
         {
             mst[v] = -;
             ans += lowcost[v];
             for(int j = ; j <= n; j++)
             {
                 if(mst[j] != - && lowcost[j] > Map[v][j])
                 {
                     lowcost[j] = Map[v][j];
                     mst[j] = v;
                 }
             }
         }
     }
     printf("%.2f\n", 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 = ; i <= n; i++)
             for(int j = ; j <= n; j++)Map[i][j] = INF * 1.0;
         memset(lowcost, , sizeof(lowcost));
         for(int i = ; i <= n; i++)
         {
             scanf("%lf%lf", &a[i].x, &a[i].y);
         }
         for(int i = ; i <= n; i++)
         {
             for(int j = i + ; j <= n; j++)
             {
                 double d = dis(a[i], a[j]);
                 Map[i][j] = Map[j][i] = d;
             }
         }
         prim();

     }
     return ;
 }