EOJ 2067 Building Roads

http://www.acm.cs.ecnu.edu.cn/problem.php?problemid=2067poj 3625 http://poj.org/problem?id=3625

 

题目大意：有N个农场，现在给出他们的坐标，并且已经知道了有些农场他们之间是已经相连的了，现在问怎么连最小的边，能将这些农场都连接起来
解题思路：变相的最小生成树，即将已经连起来的边的权值置为-1，那么根据prim算法，优先选的就是那条-1边，这样我们在选边的时候，如果是-1边我们就可以不用加进来

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

#define MAXN 1005
#define inf 1e12

struct Point
{
    double x, y;
}point[MAXN];

int n, m;
double map[MAXN][MAXN];

double dist(Point a, Point b)
{
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

void prim()
{
    int vis[MAXN], v;
    double mi, d[MAXN];
    for (int i = 1; i <= n; i++)
    {
        vis[i] = 0;
        d[i] = map[1][i];
    }
    for (int i = 1; i <= n; i++)
    {
        mi = inf;
        for (int j = 1; j <= n; j++)
            if (!vis[j] && d[j] < mi)
            {
                mi = d[j];
                v = j;
            }
        vis[v] = 1;
        for (int j = 1; j <= n; j++)
            if (!vis[j] && d[j] > map[v][j])
                d[j] = map[v][j];
    }
    d[0] = 0;
    for (int i = 1; i <= n; i++)
        if (d[i] != -1) d[0] += d[i];
    printf("%.2f
", d[0]);
}

int main()
{
    int a, b;
    while (cin >> n >> m)
    {
        for (int i = 1; i <= n; i++)
            scanf("%lf%lf", &point[i].x, &point[i].y);
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                map[i][j] = map[j][i] = dist(point[i], point[j]);
        for (int i = 1; i <= m; i++)
        {
            scanf("%d%d", &a, &b);
            map[a][b] = map[b][a] = -1;
        }
        prim();
    }
    return 0;
}