[DP]poj1018 Communication System

最新推荐文章于 2018-08-11 10:37:53 发布

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本文介绍了一种针对通信系统的设备选型优化方法，通过动态规划算法实现对不同制造商提供的设备进行选择，以达到整体系统最大性价比的目标。输入包含设备数量及各设备的制造商选项，输出为最优配置下的带宽价格比。

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Communication System

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 26958		Accepted: 9612

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

思路：

动态规划，状态转移方程f[i][j] = min{f[i-1][min(b[j],k)] + p[k]};

表示前i个设备j带宽的最小价格，(从0到所给带宽最大值)枚举带宽

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
/*
f[i][j] = min{f[i-1][min(b[j],k)] + p[k]};
枚举带宽
*/
const int inf = 1<<29;
int f[110][1100];//前i个设备j带宽的最小价格
int n;
void dp()
{
    int i, j, k;
    int m, B, P;
    int b[110], p[110];
    for(i = 1; i <= n; i++)
    {
        scanf("%d", &m);
        for(j = 1; j <= m; j++)
        {
            scanf("%d%d", &B, &P);
            b[j] = B;
            p[j] = P;
        }
        for(j = 1; j <= m; j++)
        {
            if(i == 1)
                f[1][b[j]] = min(f[1][b[j]], p[j]);
            else
            {
                for(k = 0; k < 1100; k++)
                {
                    if(f[i-1][k] != inf)
                    {
                        if(b[j] <= k)
                            f[i][b[j]] = min(f[i][b[j]], f[i-1][k] + p[j]);
                        else
                            f[i][k] = min(f[i][k], f[i-1][k] + p[j]);
                    }
                }
            }
        }
    }
}
int main()
{
    int t;
    int i, j;


    scanf("%d", &t);
    while(t-->0)
    {
        scanf("%d", &n);
        for(i = 1; i <= n; i++)
        for(j = 0; j <= 1100; j++)
            f[i][j] = inf;
        dp();
        double ans = 0, num;//ans初始化最小值，服了自己了……
        for(j = 0; j <= 1100; j++)
            if(f[n][j] != inf)
            {
                num = j*1.0 / f[n][j];
                ans = max(ans, num);
            }
        printf("%.3lf\n", ans);
    }
    return 0;
}