HDU2955 Robberies————01背包变形

系统管理员 2022-05-17 07:35 157阅读 0赞

Robberies  
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 30946 Accepted Submission(s): 11257

**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.

![这里写图片描述][con211-1010-1.jpg]

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

--------------------

这道题要以逃跑概率为dp对象，下标代表能够抢劫到的最大的金钱

#include<bits/stdc++.h>
    using namespace std;
    const int MAXN=105;
    double dp[100005];
    int v[MAXN];
    double w[MAXN];
    double p;
    int t,n;
    int main()
    {
        scanf("%d",&t);
        while(t--)
        {
            scanf("%lf %d",&p,&n);
            p=1.0-p;//逃跑概率 
            int sum=0;
            for(int i=0;i<n;i++)
            {
                scanf("%d %lf",&v[i],&w[i]);
                sum+=v[i];
                w[i]=1.0-w[i];//抢劫这个银行逃跑的概率 
            }
            memset(dp,0,sizeof(dp)); 
            dp[0]=1;//记录的是逃跑概率 
            for(int i=0;i<n;i++)
                for(int j=sum;j>=v[i];j--)
                    dp[j]=max(dp[j],dp[j-v[i]]*w[i]);
            for(int i=sum;i>=0;i--)
                if(dp[i]>=p)    
                {
                    printf("%d\n",i);
                    break;
                }
        }
        return 0;
    }

[con211-1010-1.jpg]: /images/20220517/8a245fa46e604e84a8ae277ee24fa93d.png