LightOJ

题目链接：https://vjudge.net/problem/LightOJ-1079

1079 - Just another Robbery

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

题意：

有n家银行，xx准备打劫银行。每一家银行都有其价值以及被抓概率。在被抓概率不大于P的情况下，打劫那些银行收获最大？打劫每一家银行被抓的事件相互独立。

题解：

1.设dp[i]为收获i元时最小的被抓概率。

2.由于事件独立，即：P(AB) = P(A)P(B)，因此：P(A∪B) = P(A) + P(B) - P(AB) = P(A) + P(B) - P(A)P(B) 。

3.根据第2点，可直接背包求解。

代码一：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <cmath>
 7 #include <queue>
 8 #include <stack>
 9 #include <map>
10 #include <string>
11 #include <set>
12 using namespace std;
13 typedef long long LL;
14 const int INF = 2e9;
15 const LL LNF = 9e18;
16 const int MOD = 1e9+7;
17 const int MAXN = 1e4+100;
18 
19 double dp[MAXN];
20 int main()
21 {
22     int T, n, kase = 0;
23     scanf("%d", &T);
24     while(T--)
25     {
26         double P;
27         scanf("%lf%d", &P, &n);
28         for(int j = MAXN-1; j>=1; j--) dp[j] = 1;
29         dp[0] = 0;
30         for(int i = 1; i<=n; i++)
31         {
32             int val; double pa;
33             scanf("%d%lf", &val,&pa);
34             for(int j = MAXN-1; j>=val; j--)
35                 dp[j] = min(dp[j], dp[j-val]+pa-dp[j-val]*pa);
36         }
37 
38         int k;
39         for(k = MAXN-1; k>=0; k--)
40             if(dp[k]<=P) break;
41         printf("Case %d: %d
", ++kase, k);
42     }
43 }

View Code

代码二：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <cmath>
 7 #include <queue>
 8 #include <stack>
 9 #include <map>
10 #include <string>
11 #include <set>
12 using namespace std;
13 typedef long long LL;
14 const int INF = 2e9;
15 const LL LNF = 9e18;
16 const int MOD = 1e9+7;
17 const int MAXN = 1e4+100;
18 
19 double dp[MAXN];
20 int main()
21 {
22     int T, n, kase = 0;
23     scanf("%d", &T);
24     while(T--)
25     {
26         double P;
27         scanf("%lf%d", &P, &n);
28         memset(dp, 0, sizeof(dp));
29         dp[0] = 1;
30         for(int i = 1; i<=n; i++)
31         {
32             int val; double pa;
33             scanf("%d%lf", &val,&pa);
34             for(int j = MAXN-1; j>=val; j--)
35                 dp[j] = max(dp[j], dp[j-val]*(1-pa));
36         }
37 
38         int k;
39         for(k = MAXN-1; k>=0; k--)
40             if(1-dp[k]<=P) break;
41         printf("Case %d: %d
", ++kase, k);
42     }
43 }

View Code