light oj 1079 - Just another Robbery 【01背包】

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发布时间：2019-06-22

本文共 2747 字，大约阅读时间需要 9 分钟。

Time Limit: 4 second(s)

Memory Limit: 32 MB

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

Output for Sample Input

0.04 3

1 0.02

2 0.03

3 0.05

0.06 3

2 0.03

3 0.05

0.10 3

1 0.03

2 0.02

3 0.05

Case 1: 2

Case 2: 4

Case 3: 6

Note

For the first case, if he wants to rob bank 1 and 2, then the probability of getting caught is 0.02 + (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：

哈利三人组要抢银行(⊙﹏⊙)b（老邓的教育没成功）。。。每个银行有M百万，风险是P，哈利每次抢银行的风险有一个上限，求能抢到的最多的金额。

思路：

用01背包解决问题。dp[i]记录抢劫金额为i时候可以成功逃跑的概率，最后把最大的金额找出来就可以了。

#include #include 
       
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          #include 
          
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             #include 
             
              #include 
              
               #include 
               
                #include 
                
                 #include 
                 
                  //#include 
                  
                   //#define LOACL#define space " "using namespace std;//typedef long long LL;typedef __int64 Int;typedef pair
                   
                     paii;const int INF = 0x3f3f3f3f;const double ESP = 1e-5;const double PI = acos(-1.0);const int MOD = 1e9 + 7;const int MAXN = 100 + 10;double dp[10000 + 10], ris[MAXN], p;int val[MAXN], n;int main() { int T; int Kcase = 0; scanf("%d", &T); while (T--) { scanf("%lf%d", &p, &n); int sum = 0, ans = 1; for (int i = 0; i < n; i++) { scanf("%d%lf", &val[i], &ris[i]); sum += val[i]; } memset(dp, 0, sizeof(dp)); dp[0] = 1; for (int i = 0; i < n; i++) { for (int j = sum; j >= val[i]; j--) { dp[j] = max(dp[j], dp[j - val[i]]*(1 - ris[i])); } } for(int i = sum; i >= 0;i--) if(dp[i] > 1 - p || fabs(dp[i] - 1 - p) < ESP){ ans = i; break; } printf("Case %d: %d\n", ++Kcase, ans); } return 0;}

转载于:https://www.cnblogs.com/cniwoq/p/6770785.html

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