poj 1018

Communication System

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20771		Accepted: 7361

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

网上说这题是DP，就做着看看，结果不知道怎么个dp法，看来dp思想还是没到位啊。上网搜报告，大多数是搜索+优化。dp做的也有，看了下思路，自己摸索着也搞出来了。

题意：

t组数据

第一行n，代表设备数量

每种设备有m个厂商提供。

要求每种设备选上1个，然后这中间最小的带宽除以你选择的总价，要求这个值最大。

最关键的是思路啊，数组怎么设，影响整个题目啊，这里p[i][j]表示到第i中设备、最小带宽是j的最小价格。

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 1<<28
using namespace std;
int p[105][1005];
int n,t,m;
int a,b,s;
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        for(int i=0;i<=100;i++)
        {
            for(int j=0;j<=1000;j++)p[i][j]=maxn;
        }
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&m);
            for(int j=0;j<m;j++)
            {
                scanf("%d%d",&b,&s);
                if(i==1)
                {
                    if(p[1][b]==maxn)p[1][b]=s;
                    else
                    {
                        if(p[1][b]>s)p[1][b]=s;
                    }
                }
                else
                {
                    for(int k=1;k<=1000;k++)
                    if(p[i-1][k]!=maxn)
                    {
                        if(k<b)
                        {
                            if(p[i][k]==maxn)p[i][k]=p[i-1][k]+s;
                            else if(p[i][k]>p[i-1][k]+s)p[i][k]=p[i-1][k]+s;
                        }
                        else
                        {
                            if(p[i][b]==maxn)p[i][b]=p[i-1][k]+s;
                            else if(p[i][b]>p[i-1][k]+s)p[i][b]=p[i-1][k]+s;
                        }
                    }
                }
            }
        }
        double ma=0;
        for(int i=1;i<=1000;i++)
        {
            ma=max(ma,(double)i/(double)p[n][i]);
        }
        printf("%.3f\n",ma);
    }
    return 0;
}