HDU-4463 Outlets 最小生成树

题意：就是最小生成树，题中给定了一条必须要连接的边，排序的时候不把他考虑进去即可。

代码如下：

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

int N;

struct Point {
    int x, y;
};

struct Edge {
    int a, b;
    double dis;
    bool operator < (const Edge &other) const {
        return dis < other.dis;    
    }
};

Point p[55];
Edge e[10000];
int idx;
int set[55];

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

int find(int x) {
    return set[x] = x == set[x] ? x : find(set[x]);
}

void merge(int a, int b) {
    set[a] = b;
}

int main() {
    int A, B;
    while (scanf("%d", &N), N) {
        idx = 1;
        scanf("%d %d", &A, &B);
        for (int i = 1; i <= N; ++i) {
            set[i] = i;
            scanf("%d %d", &p[i].x, &p[i].y);
        }
        for (int i = 1; i <= N; ++i) {
            for (int j = i+1; j <= N; ++j) {
                e[idx].a = i, e[idx].b = j;
                e[idx++].dis = dist(p[i], p[j]);
            }
        }
        double ret = 0;
        e[0].a = A, e[0].b = B, e[0].dis = dist(p[A], p[B]);
        sort(e+1, e+idx);
        for (int i = 0; i < idx; ++i) {
            int a = find(e[i].a), b = find(e[i].b);
            if (a != b) {
                ret += e[i].dis;
                merge(a, b);
            }
        }
        printf("%.2f\n", ret);
    }
    return 0;    
}