LightOJ 1079 Just another Robbery 概率背包

Description

 

As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He wants to make a calculated risk, and grab as much money as possible. But his friends - Hermione and Ron have decided upon a tolerable probability P of getting caught. They feel that he is safe enough if the banks he robs together give a probability less than P.

Input

 

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a real number P, the probability Harry needs to be below, and an integer N (0 < N ≤ 100), the number of banks he has plans for. Then follow N lines, where line j gives an integer M_j (0 < M_j ≤ 100) and a real number P_j . Bank j contains M_j millions, and the probability of getting caught from robbing it is P_j. A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

Output

 

For each case, print the case number and the maximum number of millions he can expect to get while the probability of getting caught is less than P.

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

 

Case 1: 2

Case 2: 4

Case 3: 6

 题意：

　　给你p,n

　　n个银行，给出每个银行可以抢的钱和 被抓的概率

　　问你在被抢概率才不超过p情况下能最多抢多少钱

题解：

　　01背包

　　设定dp[i][j] 前i个抢j元的 最小被抓概率

　　dp[i][j] = min(dp[i][j], dp[i-1][j-x] + (1-dp[i-1][j-x])*v[i]);

　

#include <cstdio>
#include <cstring>
#include <vector>
#include<iostream>
#include <algorithm>
using namespace std;
const int  N = 1e2 + 10, M = 1e4 , mod = 1e9 + 7, inf = 2e9;
int T,n,x;
double dp[M+20],f,p;
int main()
{
    int cas = 1;
    scanf("%d",&T);
    while(T--) {
        for(int i=0;i<=M;i++) dp[i] = -1;
        dp[0] = 0;
        scanf("%lf%d",&p,&n);
        for(int i=1;i<=n;i++) {
            scanf("%d%lf",&x,&f);
            for(int j=M;j>=x;j--) {
                if(dp[j-x]==-1) continue;
                ///cout<<j<<endl;
                if(dp[j]==-1) dp[j] = dp[j-x] + (1.0-dp[j-x])*f;
                else dp[j] = min(dp[j],dp[j-x] + (1.0-dp[j-x])*f);
            }
        }
        int ans = 0;///cout<<dp[M]<<endl;
        for(int i=0;i<=M;i++) {
            if(dp[i]!=-1&&dp[i] <= p) ans = i;
        }
        printf("Case %d: %d
",cas++,ans);
    }
}