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题目链接：

pid=2955">http://acm.hdu.edu.cn/showproblem.php?pid=2955

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.

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

2

4

6

题意：

抢银行。抢每一个银行被抓的几率为caught[]，互为独立事件，在容忍上限内，抢最多的钱。

01背包，但须要改变一点。须要将能抢来的最多的钱最为背包容量

代码例如以下：

/*此题关键是找到背包的容量和价值*/

#include <cstdio>

#include <cstring>

#define N 10047

int M[N];

double P[N],f[N];

double max(double a,double b)

{

	if(a > b)

		return a;

	return b;

}

int main()

{

	double p;

	int T,n,i,j,sum;

	scanf("%d",&T);

	while(T--)

	{

		sum = 0;

		memset(f,0,sizeof(f));//要对数组进行初始化

		scanf("%lf%d",&p,&n);

		p =1-p;;//成功逃走的概率

		f[0] = 1;//未抢劫一分钱那么逃走的概率就为1

		for(i = 0 ; i < n ; i++)

		{

			scanf("%d%lf",&M[i],&P[i]);

			P[i] = 1-P[i];

			sum+=M[i];//算出银行总钱数

		}

		for(i = 0 ; i < n ; i++)

		{

			for(j = sum ; j >= M[i] ; j--)

			{

				f[j] = max(f[j],f[j-M[i]]*P[i]);

			}//这里的每次成功的概率须要进行乘法运算。由于是两次成功的概率所以是乘法

		}

		for(i = sum ; i >= 0 ; i--)

		{

			if(f[i] >= p)

				break;//找出能够成功逃走而且逃走概率不超过警戒值

		}

		printf("%d\n",i);

	}

	return 0;

}