ACM: 三分法 poj 1018  (学习到三…

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这篇博客详细介绍了如何使用三分法解决ACM竞赛中的POJ 1018题目，通过实例解析和代码展示，帮助读者理解并掌握这种算法的应用。

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Communication System

Description

We have received an order from Pizoor Communications Inc. for a special communication system. The system consists of several devices. For each device, we are free to choose from several manufacturers. Same devices from two manufacturers differ in their maximum bandwidths and prices.
By overall bandwidth (B) we mean the minimum of the bandwidths of the chosen devices in the communication system and the total price (P) is the sum of the prices of all chosen devices. Our goal is to choose a manufacturer for each device to maximize B/P.

Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by the input data for each test case. Each test case starts with a line containing a single integer n (1 ≤ n ≤ 100), the number of devices in the communication system, followed by n lines in the following format: the i-th line (1 ≤ i ≤ n) starts with mi (1 ≤ mi ≤ 100), the number of manufacturers for the i-th device, followed by mi pairs of positive integers in the same line, each indicating the bandwidth and the price of the device respectively, corresponding to a manufacturer.

Output

Your program should produce a single line for each test case containing a single number which is the maximum possible B/P for the test case. Round the numbers in the output to 3 digits after decimal point.

Sample Input

1 

3


3 100 25 150 35 80 25


2 120 80 155 40


2 100 100 120 110

Sample Output

0.649



题意: 不同设备中各取出一种设备,使得这些设备带宽的最小值和它们价值的总和的比最大.



解题思路:

         1. 网上学习的三分法.



代码:

#include <cstdio>

#include <iostream>

#include <cstring>

#include <cstdlib>

using namespace std;

#define MAX 105

const int INF = (1<<29);



int n;

int b[MAX][MAX], price[MAX][MAX], count[MAX];

int all[MAX*MAX];

int num;



int cmp(const void *a, const void *b)

{

    return *(int*)a - *(int*)b;

}



inline int min(int a,int b)

{    

    return a < b ? a : b;

}



inline double max(double a, double b)

{

    return a > b ? a : b;

}



double solve(int x)

{

    double sum = 0;

    for(int i = 1; i <= n; ++i)

    {

        int mint = INF;

        for(int j = 1; j <= count[i]; ++j)

        {

            if(all[x] <= b[i][j])

                mint = min(mint, price[i][j]);

        }

        sum += mint;

    }

    return all[x] / sum;

}



int main()

{

//    freopen("input.txt","r",stdin);

    int caseNum;

    scanf("%d",&caseNum);

    while(caseNum--)

    {

        scanf("%d",&n);

        num = 1;

        for(int i = 1; i <= n; ++i)

        {

            scanf("%d",&count[i]);

            for(int j = 1; j <= count[i]; ++j)

            {

                scanf("%d %d",&b[i][j], &price[i][j]);

                all[num++] = b[i][j];

            }

        }

        qsort(all+1,num,sizeof(all[0]),cmp);

        

        int left = 1, right = num;

        int mid1, mid2;

        while(left+1 < right)

        {

            mid1 = (left+right)/2;

            mid2 = (mid1+right)/2;

            double temp1 = solve(mid1);

            double temp2 = solve(mid2);

            if(temp1 < temp2) left = mid1;

            else right = mid2;

        }

        double result = max(solve(left), solve(right));

        printf("%.3f\n",result);

    }

    return 0;

}