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Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11816    Accepted Submission(s): 4397

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

问题的关键是把被抓的概率转换成安全的概率

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
int main()
{

    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,c[110],i,j;
        double f,dp[11000]={0},w[110];
        scanf("%lf%d",&f,&n);
        f=1-f;
        for(i=0;i<n;i++)
           {
               scanf("%d%lf",c+i,w+i);
               w[i]=1-w[i];
           }
           dp[0]=1;
        for(i=0;i<n;i++)
        {
            for(j=10000;j>=c[i];j--)
            {
                if(dp[j]<dp[j-c[i]]*w[i])
                    dp[j]=dp[j-c[i]]*w[i];
            }
        }
        for(i=10000;i>=0;i--)
        {
            if(f<dp[i])
            {
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}

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HDU2955Robberies

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原文地址：http://blog.csdn.net/fljssj/article/details/38487153