Hdu 2955 Robberies（01背包）

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)
Total Submission(s): 16775    Accepted Submission(s): 6187

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.

 

题意:

给一个最大风险值（背包容量），有很多银行，Mj表示第j个银行有多少钱，Pj表示抢这个银行的风险值，求在给定最大风险值的范围内最多能抢到的金额

题解：

不难发现这是一个01背包的水题，由于代价是浮点型的，为了避免精度问题可以将问题转化为求抢劫一定金额后最大的逃跑成功的概率

#include<stdio.h>
#include<string.h>
const int N=101;
float p[N],dp[N*100],pn;
int v[N],sum;
int main()
{
    int T,n,res;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%f%d",&pn,&n);sum=0;
        pn=1-pn;//由于pn是可以承担的最大风险值 则1-pn为最低要达到的逃跑成功率 
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;++i)
        scanf("%d%f",&v[i],&p[i]),sum+=v[i],p[i]=1-p[i];
        dp[0]=1;//不抢劫的时候逃跑成功率为1 
        for(int i=0;i<n;++i)
        for(int j=sum;j>=v[i];--j)
        if(dp[j-v[i]]*p[i]>dp[j])//成功率是乘法计算的 
		dp[j]=dp[j-v[i]]*p[i];
        res=0;
        for(int i=sum;i>=0;--i)
        if(dp[i]>=pn)//只要成功率大于pn i即为要求的答案 
        {
            res=i;
            break;
        }
        printf("%d\n",res);
    }
}