Desert King(POJ2728+最优比率生成树+二分）

题目链接：http://poj.org/problem?id=2728

题目：

题意：求一颗生成树，使得费用与距离的比值最小，其中距离等于两点之间的平面欧拉距离，费用为z坐标之差。

思路：

　　由上图我们可以得知，我们只需对x进行二分（最大化平均值），以cost[i]-len[i]*x为边权跑prime即可。

代码实现如下：

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;;
typedef pair<int, int> pii;
typedef unsigned long long ull;

#define lson i<<1
#define rson i<<1|1
#define bug printf("*********
");
#define FIN freopen("D://code//in.txt", "r", stdin);
#define debug(x) cout<<"["<<x<<"]" <<endl;
#define IO ios::sync_with_stdio(false),cin.tie(0);

const double eps = 1e-8;
const int mod = 10007;
const int maxn = 1000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f;

int n, x, y, z;
double ans;
int vis[maxn];
double mp[maxn][maxn], dis[maxn];

struct node {
    int x, y ,z;
}p[maxn];

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

bool prime(double x) {
    for(int i = 1; i <= n; i++) {
        dis[i] = inf;
        vis[i] = 0;
    }
    ans = 0;
    dis[1] = 0;
    for(int i = 1; ; i++) {
        double mx = inf;
        int t = -1;
        for(int j = 1; j <= n; j++) {
            if(!vis[j] & mx > dis[j]) {
                mx = dis[j];
                t = j;
            }
        }
        if(t == -1) break;
        ans += mx;
        vis[t] = 1;
        for(int j = 1; j <= n; j++) {
            if(!vis[j] && dis[j] > fabs(p[t].z - p[j].z) - mp[t][j] * x) {
                dis[j] = fabs(p[t].z - p[j].z) - mp[t][j] * x;
            }
        }
    }
    return ans >= 0;
}

int main() {
    //FIN;
    while(~scanf("%d", &n) && n) {
        for(int i = 1; i <= n; i++) {
            scanf("%d%d%d", &p[i].x, &p[i].y, &p[i].z);
        }
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                mp[i][j] = dist(p[i], p[j]);
            }
        }
        double ub = 100, lb = 0, mid;
        while(ub - lb > eps) {
            mid = (ub + lb) / 2;
            if(prime(mid)) lb = mid;
            else ub = mid;
        }
        printf("%.3f
", lb);
    }
    return 0;
}