【MST-Prime】POJ 2560 Freckles

POJ 2560 Freckles

• 给出所有点的坐标，然后让求将这些点连起来需要的最少的路径长度。
• 先将每两点之间的距离算出来存起来，然后再用Prime求最小生成树就好了。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <list>
#include <map>
#define INF 0x3f3f3f3f
#define MID l + r >> 1
#define lsn rt << 1
#define rsn rt << 1 | 1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr

using namespace std;
typedef long long ll;
const int maxN = 100 + 5;
const int maxM = 2e5 + 5;

int n;
struct node{
    double x, y;
}coord[maxN];
double mp[maxN][maxN];
double dis[maxN];
int vis[maxN];

double Prime()
{
    double ans = 0;
    for(int i = 0; i < n ; i ++ )
    {
        dis[i] = (i == 0 ? 0 : INF);
        vis[i] = 0;
    }
    for(int j = 0; j < n ; j ++ )
    {
        double minn = INF; int min_pos = -1;
        for(int i = 0; i < n ; i ++ )
        {
            if(!vis[i] && dis[i] < minn)
            {
                minn = dis[i];
                min_pos = i;
            }
        }
        vis[min_pos] = 1;
        for(int i = 0; i < n ; i ++ )
        {
            if(dis[i] > mp[min_pos][i])
                dis[i] = mp[min_pos][i];
        }
        ans += minn;
    }
    return ans;
}

int main()
{
    while(~scanf("%d", &n))
    {
        for(int i = 0; i < n ; i ++ )
            scanf("%lf%lf", &coord[i].x, &coord[i].y);
        for(int i = 0; i < n ; i ++ )
        {
            for(int j = i + 1; j < n ; j ++ )
            {
                mp[i][j] = sqrt((coord[i].x - coord[j].x) * (coord[i].x - coord[j].x) + (coord[i].y - coord[j].y) * (coord[i].y - coord[j].y));
                mp[j][i] = mp[i][j];
            }
        }
        printf("%.2f\n", Prime());
    }
    return 0;
}