Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31769    Accepted Submission(s): 11527

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

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

题目大意：

Roy是一个盗贼。现在有n个银行成了他的下手目标。对第i个银行下手，他会获得mi元钱，同时有pi的概率被抓。问在被抓概率不超过p的情况下，他最多能得到多少钱。

01背包好题。

这题仔细想想还是很有趣的。获得的钱其实是背包容量。获得这些钱安全的概率才是dp存储的东西。要好好想想。

此外，对double类型变量输入输出是 printf&%f  scanf&%lf，注意。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<stack>

using namespace std;

const int maxn=100;

int money[maxn+5];
double safe[maxn+5];//不被抓的概率作为重量
double dp[maxn*maxn+5];//获得i钱，不被抓的概率

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        double p;int n;
        scanf("%lf%d",&p,&n);
        p=1-p;
        int sum=0;//最多获得多少钱
        for(int i=1;i<=n;i++)
        {
            scanf("%d%lf",money+i,safe+i);
            safe[i]=1-safe[i];
            sum+=money[i];
        }

        for(int i=0;i<=sum;i++)
            dp[i]=0;
        dp[0]=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=sum;j>=money[i];j--)
            {
                dp[j]=max(dp[j],dp[j-money[i]]*safe[i]);
            }
        }

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

原文地址：https://www.cnblogs.com/acboyty/p/9769881.html