Hdu 2955 Robberies 0/1背包

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10526    Accepted Submission(s): 3868

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

 

　　这是一道关于01背包的问题，题目的意思是有个强盗想去抢劫银行，但是又不想被抓到，所以，他要计算不被抓到的情况下可以获得的最大的金钱数目。首先给定一个数T表示有T组测试数据，然后是两个数P和N，P表示被抓的几率，N表示有三家银行，接下来N行是每家银行抢到的金钱和被抓的几率，输出不被抓的情况下可以抢到的最大的金额。

　　首先我们可以算出不被抓的几率和最多可以抢到的金钱，然后在这种情况下相当于01背包问题。不过要注意的是状态转移方程是dp[i] = max(dp[i],dp[i-m[i]]*p[i])，而不是dp[i] = max(dp[i],dp[i-m[i]]+p[i])，这一点相当重要，也是解题的关键。

 

提供参考代码:

#include <iostream>
#include <cstdio>
using namespace std;
#define MAX 10003
double p[MAX],f[MAX];
int m[103];
int main()
{
    double P;
    int T, N, i, j;
    cin>>T;
    while(T--)
    {
        int sum = 0;
        scanf("%lf %d",&P,&N);
        P = 1-P;    //不被抓的概率
        for(i=0; i<N; i++)
        {
            scanf("%d %lf",&m[i],&p[i]);
            p[i] = 1-p[i];  //不被抓的概率
            sum += m[i]; //可以抢到的最大金钱数目
        }
        
        for(i=0; i<=sum; i++)
            f[i] = 0;
        f[0] = 1;     //表示抢金钱为0的时候,不被抓的概率为1
        for(i=0; i<N; i++)
            for(j=sum; j>=m[i]; j--)
                f[j] = max(f[j],f[j-m[i]]*p[i]);
        for(i=sum; i>=0; i--)  //从最大的金钱数目开始，依次查看不被抓概率是否和给定的相等
            if(f[i]-P>0.000000001)
            {
                cout<<i<<endl;
                break;
            }
    }
    return 0;
}