POJ 1018 Communication System（贪心）

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

Source

Tehran 2002, First Iran Nationwide Internet Programming Contest

 

    题目大致意思：就是要做一个系统，然后需要n种配件，每种配件由于不同的厂家生产，它的参数 B （带宽）、P（价格）不一样，现在要选取这些配件，然后在n个配件中用最小的B除以所有的P的和，问怎么选才能使得B/P最大。

    题目的意思一开始没看明白，一直以为是用所有的B除以所有的P，结果手算总是算不对，搞得我都怀疑我的算数能力了。

    弄懂题目意思以后还是很容易做出来的。

    分析：

    1.我们可以先确定一个配件，以这个配件的B为n个配件中的最大B

    2.接着选后面的n-1个配件，满足要求：所选的配件的B不能大于第一个选择的配件的B

    3.后面的n-1种器件的P应该尽可能小

    4.选好以后，求出B/P的值，比较不同的选法那个B/P最大，输出最大的那个

 

代码如下：

#include <iostream>
#include <vector>
 
using namespace std;
 
struct goods
{
    int p, b;
};
vector<goods>a[];
const int MAXN = ;
 
void Clear()                                            //用来清空信息的  很重要 没有这个会错
{
    for (int i = ; i <= ; i++)
        a[i].clear();
}
 
int main()
{
    int t, num, n;
    double ans;
    goods a1, a2;
    cin >> t;
    while (t--)
    {
        Clear();
        ans = -;
        cin >> num;         //器件总数
        for (int i = ; i < num; i++)
        {
            cin >> n;
            for (int j = ; j < n; j++)
                cin >> a1.b >> a1.p, a[i].push_back(a1);
        }
        for(int i = ; i < num; i++)                          //枚举第i个器件
        {
            for (int j = ; j < a[i].size(); j++)             //枚举第i种器件的对应器件 确定第i种器件
            {
                a1 = a[i][j];                                 //选中第i种器件的第j个  而且假定这个器件的B是最小的
                double B = a1.b;
                double P = a1.p;
                int pp;
                for (int k = ; k < num; k++)                 //枚举n-1种器件
                {
                    if (k == i) continue;                     //第i种器件已经选择过了 避免重复选择
                    pp = MAXN;
                    for (int h = ; h < a[k].size(); h++)
                    {
                        a2 = a[k][h];
                        if (a2.b < B) continue;               //假定B是最小的 不能选择更小的
                        if (pp > a2.p) pp = a2.p;             //为了使B/P尽可能大 P就要尽可能小
                    }
                    if (pp == MAXN) break;                    //找不到合适的器件 那么就表示选错了  重新选择
                    P += pp;
                }
                if (pp == MAXN) break;
                if (ans < (B / P)) ans = (B / P);
            }
        }
        printf("%.3f\n", ans);
    }
    return ;
}