Communication System_DP

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 27048		Accepted: 9641

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】要建一个通讯系统，有n种设备，每种设备有m小类，n种设备中各取一种，使得B/P最大，其中B是选取设备中最小的b，而P是选取设备p之和；

【思路】DP，要使B/P最大，即B尽量大，P尽量小

           dp[i][j]表示第i种设备，b最小为j的时候P的值为dp[i][j];

           dp[i][b]=min(dp[i][b],dp[i][k]+p) k>b;

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int inf=0x7777777;
const int N=10010;
int dp[105][N];
int main()
{
    int t,n,m,b,p;
    cin>>t;
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            for(int j=0; j<N; j++)
            {
                dp[i][j]=inf;
            }
        }
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&m);
            for(int j=1; j<=m; j++)
            {
                scanf("%d%d",&b,&p);
                if(i==1) dp[1][b]=min(dp[1][b],p);//DP
                else
                {
                    for(int k=0; k<N; k++)
                    {
                        if(k<=b)
                            dp[i][k]=min(dp[i][k],dp[i-1][k]+p);
                        else dp[i][b]=min(dp[i][b],dp[i-1][k]+p);
                    }
                }
            }
        }
        double ans=0;
        for(int i=0; i<N; i++)
        {
            if(dp[n][i]!=inf)
            {
                double tmp=(double)i/dp[n][i];
                if(tmp>ans) ans=tmp;
            }

        }
        printf("%.3f
",ans);
    }
    return 0;
}