poj2728 Desert King【最优比率生成树】【Prim】【0/1分数规划】

含【最小生成树Prim】模板。

Prim复杂度为$O(n^2),适用于稠密图，特别是完全图的最小生成树的求解。

 

Desert King

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions:31622		Accepted: 8670

Description

David the Great has just become the king of a desert country. To win the respect of his people, he decided to build channels all over his country to bring water to every village. Villages which are connected to his capital village will be watered. As the dominate ruler and the symbol of wisdom in the country, he needs to build the channels in a most elegant way. 

After days of study, he finally figured his plan out. He wanted the average cost of each mile of the channels to be minimized. In other words, the ratio of the overall cost of the channels to the total length must be minimized. He just needs to build the necessary channels to bring water to all the villages, which means there will be only one way to connect each village to the capital. 

His engineers surveyed the country and recorded the position and altitude of each village. All the channels must go straight between two villages and be built horizontally. Since every two villages are at different altitudes, they concluded that each channel between two villages needed a vertical water lifter, which can lift water up or let water flow down. The length of the channel is the horizontal distance between the two villages. The cost of the channel is the height of the lifter. You should notice that each village is at a different altitude, and different channels can't share a lifter. Channels can intersect safely and no three villages are on the same line. 

As King David's prime scientist and programmer, you are asked to find out the best solution to build the channels.

Input

There are several test cases. Each test case starts with a line containing a number N (2 <= N <= 1000), which is the number of villages. Each of the following N lines contains three integers, x, y and z (0 <= x, y < 10000, 0 <= z < 10000000). (x, y) is the position of the village and z is the altitude. The first village is the capital. A test case with N = 0 ends the input, and should not be processed.

Output

For each test case, output one line containing a decimal number, which is the minimum ratio of overall cost of the channels to the total length. This number should be rounded three digits after the decimal point.

Sample Input

4
0 0 0
0 1 1
1 1 2
1 0 3
0

Sample Output

1.000

Source

Beijing 2005

 

题意：

有$n$个城市，每个城市有对应的坐标和海拔。两个城市之间的距离是坐标的距离，路的花费是海拔之差。

现在要建$n-1$条路，使得花费之和比距离之和最小。

思路：

一道0/1分数规划的题目。

0/1分数规划模型指：给定整数$a_1,a_2,...,a_n$,以及$b_1,b_2,...,b_n$,求一组解$x_i(1\leq i \leq n,x_i=0或1)$，

使得$\frac{\sum_{i=1}^{n} a_i*x_i}{\sum_{i=1}^{n}b_i*x_i}$最大化。

这种类型的题目我们可以通过二分答案来做。

因为如果对于$mid$，存在一组解，使得$\sum_{i=1}^{n}(a_i - mid * b_i) * x_i \geq 0$那么我们可以变形得到

存在一组解，使得$\frac{\sum_{i=1}^{n} a_i*x_i}{\sum_{i=1}^{n}b_i*x_i} \geq mid$

也就是说，$mid$比我们实际的答案要小。

反之，如果对于任意一组解，都有$\sum_{i=1}^{n}(a_i - mid * b_i) * x_i <0$那么我们可以得到

任意的解都有$\frac{\sum_{i=1}^{n} a_i*x_i}{\sum_{i=1}^{n}b_i*x_i} < mid$

也就是说，$mid$比我们实际的答案要大。

所以我们每次只需要计算$\sum_{i=1}^{n}(a_i - mid * b_i)*x_i$的最大值，如果最大值非负，令$st = mid$，否则$ed = mid$

 

对于这道题，也是类似。我们二分最终的答案。

每一次重新建图，城市和城市之间的边变为$cost - mid * length$

然后在新图上跑最小生成树。

如果最小生成树的值非负，说明实际答案比$mid$要大，令$st = mid$

由于这道题是完全图，而点的个数只有$1000$所以用prim比较好

 

WA了好久，后来发现是对d赋初值的问题。

对于double的数组赋初值$+\infty$不能用$0x3f$而应该用$0x7f$，就改了这里就过了。

#include<iostream>
//#include<bits/stdc++.h>
#include<cstdio>
#include<cmath>
//#include<cstdlib>
#include<cstring>
#include<algorithm>
//#include<queue>
#include<vector>
//#include<set>
//#include<climits>
//#include<map>
using namespace std;
typedef long long LL;
#define N 100010
#define pi 3.1415926535
#define inf 0x3f3f3f3f

const int maxn = 1005;
const double eps = 1e-7;
int n;
struct city{
    double x, y, height;
}c[maxn];
double dist[maxn][maxn], cost[maxn][maxn], g[maxn][maxn];

double get_dist(int i, int j)
{
    return sqrt((c[i].x - c[j].x) * (c[i].x - c[j].x) + (c[i].y - c[j].y) * (c[i].y - c[j].y));
}


double d[maxn];
bool vis[maxn];
void prim()
{
    memset(d, 0x7f, sizeof(d));
    memset(vis, 0, sizeof(vis));
    /*for(int i = 1; i <= n; i++){
        d[i] = (double)inf;
        vis[i] = 0;
    }*/
    d[1] = 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] = 1;
        for(int y = 1; y <= n; y++){
            if(!vis[y])d[y] = min(d[y], g[x][y]);
        }
    }
}

void getgraph(double mid)
{
    for(int i = 1; i <= n; i++){
        for(int j = i; j <= n; j++){
            g[i][j] = g[j][i] = cost[i][j] - mid * dist[i][j];
        }
    }
}

bool check(double mid)
{
    getgraph(mid);
    prim();
    double res = 0;
    for(int i = 1; i <= n; i++){
        res += d[i];
    }
    return res >= 0;
}


int main()
{
    while(scanf("%d", &n) != EOF && n){
        for(int i = 1; i <= n; i++){
            scanf("%lf%lf%lf", &c[i].x, &c[i].y, &c[i].height);
        }

        for(int i = 1; i <= n; i++){
            for(int j = i; j <= n; j++){
                dist[i][j] = dist[j][i] = get_dist(i, j);
                cost[i][j] = cost[j][i] = fabs(c[i].height - c[j].height);
            }
        }
        double st = 0.0, ed = 100.0;
        while(ed - st >= eps){
            double mid = (st + ed) / 2;
            if(check(mid))st = mid;
            else ed = mid;
        }
        printf("%.3f\n", ed);
    }

    return 0;
}