POJ 2728 Desert King (算竞进阶习题)

最小生成树 + 01分数规划

好烦的卡精度题，调了一下午。。

题意大概就是给你一张稠密图，每条边都有收益和花费，要求一个花费之和与收益之和的比值和最小的生成树。

当然我们可以看成收益之和与花费之和的比值最大的生成树。。

这显然是一个01分数规划问题。然后二分枚举猜测的L值跑最大生成树就行了，若最大生成树大于等于0，代表我们的最大值大于等于L，于是mid = l，若最大生成树小于0，代表我们的最大值小于L， 于是mid = r。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iomanip>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 1005;
const double eps = 1e-7;
int n;
double x[N], y[N], z[N];
double dist[N][N], cost[N][N], cur[N][N], mst, d[N];
bool vis[N];

inline double getDis(double x1, double y1, double x2, double y2){
    return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}

inline void build(double mid){
    for(int i = 1; i <= n; i ++){
        for(int j = 1; j <= n; j ++) cur[i][j] = -INF;
    }
    for(int i = 1; i <= n; i ++){
        for(int j = 1; j <= n; j ++){
            if(i == j) continue;
            cur[i][j] = dist[i][j] - mid * cost[i][j];
        }
    }
}

inline void prim(){
    for(int i = 1; i <= n; i ++){
        d[i] = -INF;
        vis[i] = false;
    }
    d[1] = mst = 0.0;
    for(int i = 1; i < n; i ++){
        int x = 0;
        for(int j = 1; j <= n; j ++){
            if(!vis[j] && (x == 0 || d[j] > d[x])) x = j;
        }
        vis[x] = true;
        for(int j = 1; j <= n; j ++){
            if(!vis[j]) d[j] = max(d[j], cur[x][j]);
        }
    }
    for(int i = 2; i <= n; i ++) mst += d[i];
}

inline bool check(){
    prim();
    return mst >= 0.0;
}

int main(){

    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    while(cin >> n && n){
        for(int i = 1; i <= n; i ++)
            cin >> x[i] >> y[i] >> z[i];
        for(int i = 1; i <= n; i ++){
            for(int j = i + 1; j <= n; j ++){
                dist[i][j] = dist[j][i] = getDis(x[i], y[i], x[j], y[j]);
                cost[i][j] = cost[j][i] = fabs(z[i] - z[j]);
            }
        }
        double l = 0.0, r = 40.0;
        while(r - l >= eps){
            double mid = (l + r) / 2;
            build(mid);
            if(check()) l = mid;
            else r = mid;
        }
        printf("%.3lf\n", 1 / l);
    }
    return 0;
}