LightOJ 1079 概率型 01背包

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 thanP.

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 integerN (0 < N ≤ 100), the number of banks he has plans for. Then followN lines, where line j gives an integer M_j (0 < M_j ≤ 100) and a real number P_j . Bankj contains M_j millions, and the probability of getting caught from robbing it isP_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 thanP.

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

Hint

For the first case, if he wants to rob bank 1 and 2, then the probability of getting caught is0.02 + (1 - 0.02) * .03 = 0.0494 which is greater than the given probability(0.04). That's why he has only option, just to rob rank 2.

题意：我想抢劫银行，当危险率低于P的时候才能行动，现在给出每家银行的金钱mi和危险率pi，求最多能获得多少金钱


题解：危险率是P，那么安全率就是1-P，那么我抢劫的所有银行的安全率之积就不能小于1-P，这样就变成了一个01背包

#include<math.h>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define MAXN 110#define eps 1e-6struct bank{    int m;    double p;}a[MAXN];double dp[MAXN*MAXN];int main(){    int n,T;    int cases=1;    freopen("in.txt","r",stdin);    scanf("%d",&T);    while(T--)    {        int sum=0;        double num;        scanf("%lf%d",&num,&n);        for(int i=1;i<=n;i++){            scanf("%d%lf",&a[i].m,&a[i].p);            a[i].p=1-a[i].p;            sum+=a[i].m;        }        memset(dp,0,sizeof(dp));        dp[0]=1.0;        for(int i=1;i<=n;i++)            for(int j=sum;j>=a[i].m;j--)                dp[j]=max(dp[j],dp[j-a[i].m]*a[i].p);        int ans=0;        for(int i=0;i<=sum;i++)            if(1-dp[i]<=num)            ans=max(ans,i);        printf("Case %d: %d\n",cases++,ans);    }    return 0;}