LightOJ 1079 Just another Robbery

LightOJ 1079 Just another Robbery

题目链接

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 Mj (0 < Mj ≤ 100) and a real number Pj . Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj. 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

很明显的概率背包 DP ~
我们直接求成功的概率不好求，可以反过来，求失败的概率，用 $dp[i]$ 表示偷 $i$ 百万元的最大失败概率，最后只需遍历一遍 $1->sum$，只要满足 $1-dp[i]<1-p$ 即 $dp[i]>p$ 就可以直接输出了，背包的板子从加改为乘即可，AC代码如下：

#include
using namespace std;
typedef long long  ll;
int t,n,w[105];
double p,dp[10005],v[105],eps=1e-8;
int main(){
    scanf("%d",&t);
    int id=1;
    while(t--){
        scanf("%lf%d",&p,&n);
        p=1-p;
        int sum=0;
        for(int i=0;i<n;i++) scanf("%d%lf",&w[i],&v[i]),sum+=w[i],v[i]=1-v[i];
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(int i=0;i<n;i++){
            for(int j=sum;j>=w[i];j--){
                dp[j]=max(dp[j],dp[j-w[i]]*v[i]);
            }
        }
        for(int i=sum;i>=0;i--){
            if(dp[i]-p>eps){
                printf("Case %d: %d\n",id++,i);
                break;
            }
        }
    }
	return 0;
}