HDU - 2955 Robberies 【0-1背包】

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本文探讨了一个银行抢劫场景下的数学优化问题，通过将银行抢劫转化为背包问题，利用动态规划求解在限定被捕风险下，如何最大化抢劫金额。文章详细介绍了问题转换的思路与AC代码实现。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2955

Robberies

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Problem Description

The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usually gets caught in the end, often because they become too greedy. He has decided to work in the lucrative 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 the security of various banks and the amount of cash they hold. He wants to make a calculated risk, and grab as much money as possible.

His mother, Ola, has decided upon a tolerable probability 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 gives T, the number of cases. For each scenario, the first line of input gives a floating point number P, the probability Roy needs to be below, and an integer 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 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 if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

Sample Input

Sample Output

2
4
6

解题思路

这题很容易想成：以最大被抓概率为背包最大容量，将银行作为商品，抢一个银行的被抓概率为商品体积，银行里的钱为商品价格。这有两个问题：①被抓概率为浮点数，且不知道小数点后几位（double16位必然爆内存）无法开数组；②被抓概率不是简单的相加，（因为你第一次逃掉的概率为A,第二次逃掉的概率为B，那么两次都逃掉的概率是A*B不是A+B）。

那么我们不妨以所有银行的钱数总和为背包最大容量，以抢某个银行的钱作为商品体积，以抢完（后）这个银行后的逃跑概率作为商品价值（背包模型中数组的值代表能得到的最大价值，这里数组的值代表能跑掉的最大概率）。那么我们从小到大枚举背包容量（钱），当商品价值（逃跑概率）大于最小逃跑概率时，所对应的背包容量（钱）就是可以得到的最优解（能抢的最多钱）。

AC代码

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
struct bank {
	double w;
	int c; 
}b[101];
double escape[10005];	//当抢了i元时的逃跑概率 
int main()
{
	int T,N;
	double P;
	cin >> T;
	while( T--)
	{
		cin >> P >> N;
		int sum=0;	//所有银行的钱 
		for(int i=0;i<N;i++) {
			cin >> b[i].c >> b[i].w;
			sum += b[i].c; 
		}
		memset(escape,0,sizeof escape);	//刚开始认为只要抢了钱，你就跑不掉 
		escape[0] = 1;	//啥都没抢凭啥抓你？所以逃跑概率当然是1啦 
		for(int i=0;i<N;i++) 
			for(int j=sum;j>=b[i].c;j--)	//抢你一块钱或者把你的钱都抢了你都要抓我，对强盗来说当然要全抢才可能是最优解 
				escape[j] = max(escape[j],escape[j-b[i].c]*(1-b[i].w));	//多种抢钱方法得到的钱一样，我当然去抢逃跑概率大的那种 
		
		for(int i=sum;i>=0;i--)
			if(escape[i] > double(1-P)) {	//再多抢就跑不掉了 
				cout << i << endl;
				break;
			} 
	}
}