HDU 2955 Robberies

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17647    Accepted Submission(s): 6516

Problem Description

The aspiring(有抱負的) Roy the Robber has seen a lot of American movies, and knows that the bad guys(球員) usually gets caught in the end, often because they become too greedy. He has decided to work in the lucrative(有利可圖的) business of bank robbery only for a short while, before retiring to a comfortable job at a university.


For a few months now, Roy has been assessing(評定) the security of various banks and the amount(數量) of cash they hold. He wants to make a calculated(計算出的) risk(風險), and grab as much money as possible.


His mother, Ola, has decided upon a tolerable(可以的) probability(可能性) of getting caught. She feels that he is safe enough if the banks he robs together give a probability less than this.

 

Input

The first line of input(投入) gives T, the number of cases. For each scenario(方案), the first line of input gives a floating point number P, the probability(可能性) Roy needs to be below, and an integer(整數) N, the number of banks he has plans for. Then follow N lines, where line j gives an integer Mj and a floating point number Pj .
Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .

 

Output

For each test case, output(輸出) a line with the maximum number of millions he can expect to get while the probability of getting caught is less than the limit set.

Notes and Constraints
0 < T <= 100
0.0 <= P <= 1.0
0 < N <= 100
0 < Mj <= 100
0.0 <= Pj <= 1.0
A bank goes bankrupt(破產者) if it is robbed, and you may assume(承擔) that all probabilities are independent as the police have very low funds.

 

Sample Input

3 0.04 3 1 0.02 2 0.03 3 0.05 0.06 3 2 0.03 2 0.03 3 0.05 0.10 3 1 0.03 2 0.02 3 0.05

 

Sample Output

2 4 6

 

題目意思是給定給定一個概率，是被抓的概率，給定幾個銀行，每個銀行對應一個錢數和被抓的概率，如何在有效的被抓概率內偷更多的錢？

顯然是個0-1揹包的問題，一開始把這個當成了一個概率簡單相加的問題，看了討論區發現傻了。。其實概率應該是相乘的。設個一維DP，代表對應money被抓的概率，最後反向尋找第一個概率大於1 - 被抓的概率的就好。

代碼如下：

/*************************************************************************
    > File Name: 0-1_bags.cpp
    > Author: Zhanghaoran
    > Mail: [email protected]
    > Created Time: Fri 27 Nov 2015 11:21:43 PM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;
float dp[10010];
int main(void){
    int T, n;
    float p;
    int mon[10010];
    float pro[10010];
    int all = 0;
    cin >> T;
    while(T --){
	all = 0;
        scanf("%f%d",&p,&n);
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i ++){
            scanf("%d%f",&mon[i],&pro[i]);
            all += mon[i];
        }

        dp[0] = 1;
        for(int i = 1; i <= n; i ++){
            for(int j = all; j >= mon[i]; j --){
                dp[j] = max(dp[j], dp[j - mon[i]] * (1 - pro[i]));
            }
        }

        for(int i = all; i >= 0; i --){
            if(dp[i] >= (1 - p)){
                printf("%d\n", i);
                break;
            }
        }

    }
    return 0;

}