Just another Robbery（概率dp）

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：Harry Potter失业了，于是他想到去rob bank，对于每个bank，各自总财产一定，为v[ i ]，且Harry Potter在各个bank被抓的概率为p[ i ]。Harry Potter的盆友们认为当被抓的到的概率小于等于q的时候，Harry Potter是安全的，问当Harry Potter是安全的能够获得的最大的钱是多少。

题解：n<=100，v[ i ]<=100。然后结合01背包，可以定义dp[ i ][ 2 ]数组，dp[ i ][ 0 ]表示获得i这么多钱的时候没有被捕的概率，dp[ i ][ 0 ]表示获得i这么多钱的时候被捕的概率。

于是跑01背包。状态转移为

if(dp[j][1]>dp[j-v[i]][1]+dp[j-v[i]][0]*p[i]){
dp[j][0] = dp[j-v[i]][0]*(1-p[i]);
dp[j][1] = dp[j-v[i]][1]+dp[j-v[i]][0]*p[i];
}

代码：

#include <cstdio>#include <cstdlib>#include <iostream>#include <cstring>#include <string>#include <cstdlib>#include <algorithm>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <cmath>#include <ctime>using namespace std;typedef long long ll;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const ll LINF = 0x3f3f3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-10;const int maxn = 1e5+7;const int mod = 10007;int t,n,I = 0;double p[105],q,dp[10005][2];int v[105];int main(){    scanf("%d",&t);    while(t--){        scanf("%lf",&q);        scanf("%d",&n);        int sum = 0;        for(int i = 0;i < n;i++){            scanf("%d %lf",&v[i],&p[i]);            sum+=v[i];        }        for(int i = 0;i <= sum;i++) dp[i][0] = dp[i][1] = 1e100;        dp[0][0] = 1;        dp[0][1] = 0;        for(int i = 0;i < n;i++){            for(int j = sum;j >= v[i];j--){                if(dp[j][1]>dp[j-v[i]][1]+dp[j-v[i]][0]*p[i]){                    dp[j][0] = dp[j-v[i]][0]*(1-p[i]);                    dp[j][1] = dp[j-v[i]][1]+dp[j-v[i]][0]*p[i];                }            }        }        for(int i = sum;i >= 0;i--){            if(dp[i][1]<=q){                printf("Case %d: %d\n",++I,i);                break;            }        }    }    return 0;}