【MST】2011 regional Beijing Site A—— Qin Shi Huang's National Road System——hdu 4081

这题可以算是MST的变形，首先求MST，作用是枚举每条边。然后A = 左子树最大值 + 右子树最大值，B = 余下长度，注意枚举的那条边不是题目中的特殊边，因为如果这样想就变成了B固定，求A的最优值，而题目中A和B都是变量，互相牵制。

我在这里不是枚举边，这样做太麻烦了，改成枚举任意两点，dfs求出两点之间边权最大的边，把它替换了。

额。。。写的有点搓，各位尽情bs吧

#include <map>
#include <set>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <string>
#include <time.h>
#include <cstdio>
#include <math.h>
#include <iomanip>
#include <cstdlib>
#include <limits.h>
#include <string.h>
#include <iostream>
#include <fstream>
#include <algorithm>
using namespace std;

#define LL long long
#define MIN INT_MIN
#define MAX INT_MAX
#define PI acos(-1.0)
#define FRE freopen("input.txt","r",stdin)
#define FF freopen("output.txt","w",stdout)
#define N 1005
double max(double a, double b){return a>b?a:b;}
double dis[N];
double g[N][N];
double d[N][N];
bool vis[N];
double maxx[N][N];
vector<int> v[N];
struct node {
    double x, y;
    int num;
}p[N];
double diss(node a, node b) {
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
void dfs(int gg,int x,double maxm) {
    int i;
    vis[x] = 1;
    for(i = 0; i < v[x].size(); i++) {
        int y = v[x][i];
        if (!vis[y]) {
            maxm = max(maxm,g[x][y]);
            maxx[gg][y] = maxx[y][gg] = maxm;
            dfs(gg,y,maxm);
        }
    }
}
int main() {
    int t;
    scanf("%d",&t);
    while(t--) {
        int n;
        scanf("%d",&n);
        int i,j;
        for (i = 1;i <= n; i++) {
            scanf("%lf%lf%d",&p[i].x, &p[i].y, &p[i].num);
            v[i].clear();
        }
        memset(dis,MAX,sizeof(dis));
        for (i = 1; i <= n; i++) {
            for (j = 1; j <= n; j++) {
                d[i][j] = d[j][i] = diss(p[i], p[j]);
                g[i][j] = g[j][i] = 0;
            }
        }
        for (i = 1; i <= n; i++) {
            dis[i] = d[1][i];
            vis[i] = 0;
        }
        vis[1] = 1;
        int now = 1;
        double sum = 0;
        for (i = 1;i < n; i++) {
            double minm = MAX;
            int k = -1;
            for (j = 1; j <= n; j++) {
                if(dis[j] < minm && !vis[j]) {
                    k = j;
                    minm = dis[j];
                }
            }
            if(k != -1) {
                vis[k] = 1;
                g[k][now] = g[now][k] = minm;
                v[k].push_back(now);
                v[now].push_back(k);
                now = k;
                sum += dis[k];
                for (j = 1; j <= n; j++) {
                    if(!vis[j] && d[k][j] < dis[j]) {
                        dis[j] = d[k][j];
                    }
                }
            }
        }
        for (i = 1; i <= n; i++) {
            memset(vis,0,sizeof(vis));
            vis[i] = 1;
            int next = v[i][0];
            maxx[i][next] = g[i][next];
            maxx[next][i] = g[next][i];//就是这里wa！！！无向的！！！
            dfs(i,next,g[i][next]);
        }
        double ans = MIN;
        for (i = 1; i <= n; i++) {
            for (j = 1; j <= n; j++) {
                if(i == j)continue;
                int A = p[i].num + p[j].num;
                double res = (double)A / (sum - maxx[i][j]);
                ans = max(ans, res);
            }
        }
        printf("%.2f\n",ans);
    }
    return 0;
}