hdu2955Robberies

Robberies
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22470 Accepted Submission(s): 8307

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

小偷想偷各大银行的钱，现知道小偷的不安全系数p和银行数n
输入n行 银行存款，银行被抓系数；
求小偷处于不被抓（安全系数之内）的情况下能偷到的最多钱

01背包问题，只不过递推式有点变化，首先如果小偷不偷，那么一定安全，也就是说dp｛0｝=1;
dp[j]=max(dp[j],dp[j-a[i].val]*a[i].sfy);为递推式，表示在前j个银行中，能获取的最多钱
val代表钱，sfy代表不安全系数

因为p为不安全系数，那么安全系数就是1-p

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;struct node{    int val;    double sfy;};int main(){    node a[110];    //double dp[11000];    int n;    int t;    double p;    while(scanf("%d",&t)==1)    {        while(t--)        {            scanf("%lf%d",&p,&n);            p=1-p;            int i,j,sum=0;            for(i=0; i<n; i++)            {                scanf("%d%lf",&a[i].val,&a[i].sfy);                a[i].sfy=1-a[i].sfy;                sum+=a[i].val;            }            double dp[10010]={1.0};            for(i=0; i<n; i++)                for(j=sum; j>=a[i].val; j--)                {                    dp[j]=max(dp[j],dp[j-a[i].val]*a[i].sfy);                }            for(i=sum;i>=0;i--)            {                if(dp[i]-p>=0.000000001)                {                    cout<<i<<endl;                    break;                }            }        }    }    return 0;}