POJ 1018 Communication System

Communication System

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 28974 Accepted: 10303

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 33 100 25 150 35 80 252 120 80 155 402 100 100 120 110

Sample Output

0.649

原文链接:点击打开链接

思路： 动态规划 dp[i][j]为前i个设备且带宽为j时的最小价格和。

转移方程为： dp[i][j] = max(dp[i][j], dp[i-1][k] + p)

思路参考了 点击打开链接   这篇博客， 在原有的基础上，加了状态压缩，较少了空间消耗。

代码如下：

#include<iostream>#include<vector>#include<algorithm>#include<stdio.h>#include<limits.h>#include<memory.h>using namespace std;const int maxn = 1200;int main() {    //freopen("input.txt","r",stdin);    int t;    scanf("%d",&t);    while(t--) {        int dp[2][maxn];        for(int i = 0; i < 2; i++) {            for(int j = 0; j < maxn; j++) {                dp[i][j] = INT_MAX;            }        }        int n;        scanf("%d",&n);        int idx = 0;        for(int i = 0; i < n; i++) {            int m; scanf("%d",&m);            for(int j = 0; j < m; j++) {                int b,p;                scanf("%d%d",&b,&p);                if(0 == i) {                    dp[idx][b] = min(dp[idx][b] ,p);                }                else {                    for(int k = 0; k < maxn; k++) {                        if(dp[1-idx][k] != INT_MAX) {                            if(k <= b) {                                dp[idx][k] = min(dp[idx][k], dp[1-idx][k] + p);                            }                            else {                                dp[idx][b] = min(dp[idx][b], dp[1-idx][k] + p);                            }                        }                    }                }            }            idx = 1 - idx;            for(int k = 0; k < maxn; k++) {                dp[idx][k] = INT_MAX;            }        }        double res = 0;        idx = 1-idx;        for(int i = 0; i < maxn; i++) {            if(dp[idx][i] != INT_MAX) {                double k = double(i) / dp[idx][i];                if(k > res) res = k;            }        }        printf("%.3f\n",res);    }    return 0;}