HDU 4081 Qin Shi Huang's National Road System

HDU_4081

    这个题目大体的思路是这个样子的，首先求一个最小生成树，然后求出或者顺便求出最小生成树上任意两点间路径上的最大边权，之后枚举magic road，把magic road两个端点间的之前记录的最大边边权减去后作为B，把magic road两个端点上的人口和作为A，更新一下A/B即可。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#define MAXD 1010
#define MAXM 1000010
int N, x[MAXD], y[MAXD], people[MAXD], a[MAXM], b[MAXM], M, e, r[MAXM];
int first[MAXD], next[2 * MAXD], v[2 * MAXD], vis[MAXD], p[MAXD];
double w[MAXM], maxlen[MAXD][MAXD], total, len[2 * MAXD];
int cmp(const void *_p, const void *_q)
{
    int *p = (int *)_p;
    int *q = (int *)_q;
    return w[*p] < w[*q] ? -1 : 1;
}
int find(int x)
{
    return p[x] == x ? x : (p[x] = find(p[x]));
}
double distance(int i, int j)
{
    return sqrt((double)(x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]));
}
void init()
{
    int i, j;
    scanf("%d", &N);
    for(i = 0; i < N; i ++)
        scanf("%d%d%d", &x[i], &y[i], &people[i]);
    M = 0;
    for(i = 0; i < N; i ++)
        for(j = i + 1; j < N; j ++)
        {
            a[M] = i;
            b[M] = j;
            w[M] = distance(i, j);
            M ++;
        }
}
void add(int x, int y, double z)
{
    v[e] = y;
    len[e] = z;
    next[e] = first[x];
    first[x] = e;
    e ++;
}
void dfs(int fa, int cur, double max)
{
    int i;
    vis[cur] = 1;
    for(i = first[cur]; i != -1; i = next[i])
        if(!vis[v[i]])
        {
            if(len[i] > max)
            {
                maxlen[fa][v[i]] = len[i];
                dfs(fa, v[i], len[i]);
            }
            else
            {
                maxlen[fa][v[i]] = max;
                dfs(fa, v[i], max);
            }
        }
}
void prepare()
{
    int i, j, k, tx, ty;
    for(i = 0; i < M; i ++)
        r[i] = i;
    qsort(r, M, sizeof(r[0]), cmp);
    for(i = 0; i < N; i ++)
        p[i] = i;
    e = 0;
    memset(first, -1, sizeof(first));
    total = 0;
    for(i = 0; i < M; i ++)
    {
        k = r[i];
        tx = find(a[k]);
        ty = find(b[k]);
        if(tx != ty)
        {
            total += w[k];
            p[tx] = ty;
            add(a[k], b[k], w[k]);
            add(b[k], a[k], w[k]);
        }
    }
    for(i = 0; i < N; i ++)
    {
        memset(vis, 0, sizeof(vis));
        dfs(i, i, 0);
    }
}
void solve()
{
    int i, j;
    double temp, res = 0;
    for(i = 0; i < N; i ++)
        for(j = i + 1; j < N; j ++)
        {
            temp = (double)(people[i] + people[j]) / (total - maxlen[i][j]);
            if(temp > res)
                res = temp;
        }
    printf("%.2f\n", res);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t --)
    {
        init();
        prepare();
        solve();
    }
    return 0;
}