EOJ 3201/AOJ 2224 Save your cats【Kruskal】

題目簡介

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n個點，m條邊的圖，要使得圖中沒有圈，求要去掉的邊的權值和的最小值。

說明

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求圖的最大生成樹，用總權值減生成樹權值得到答案。似乎是隻能用kruskal做。邊數組應該是要開到10000 * 10000 / 2的，沒想到eoj上提交rte，改成10000 * 100就好了……

#include <bits/stdc++.h>
using namespace std;

#define maxn 11000

struct edge
{
    int u, v;
    double w;
}g[maxn*100];

struct node
{
    double x, y;
}point[maxn];

double sum;
int n, m, fa[maxn];

inline double dis(const node& a, const node& b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

bool cmp(const edge& a, const edge& b)
{
    return a.w > b.w;
}

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

inline double kruskal()
{
    sort(g, g+m, cmp);
    double remain = 0.0;
    for (int i = 0; i < m; ++i)
    {
        int x = find(g[i].u), y = find(g[i].v);
        if (x != y)
        {
            fa[x] = y;
            remain += g[i].w;
        }else continue;
    }
    return sum - remain;
}

int main()
{
    int u, v;
    cin >> n >> m;
    for (int i = 1; i <= n; ++i)
        fa[i] = i;
    for (int i = 1; i <= n; ++i)
        scanf("%lf%lf", &point[i].x, &point[i].y);
    for (int i = 0; i < m; ++i)
    {
        scanf("%d%d", &u, &v);
        g[i].u = u;
        g[i].v = v;
        g[i].w = dis(point[u], point[v]);
        sum += g[i].w;
    }
    printf("%.6f\n", kruskal());
    return 0;
}