題干：

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

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.

題目大意：

有n個銀行,每個銀行有v[i]的錢,和被抓住的概率p[i]。你要按一定順序去搶，要求你被抓住的概率小于等于P，你最多能獲得多少錢。?

解題報告：

首先不難證明，假設你選擇了其中某num個銀行，那么與選擇的銀行的順序無關。證明如下：（假設你選擇了第i個和的第j個）

考慮被抓的概率：，所以是等效的。

或者考慮不被抓的概率：，所以是等效的。

設定狀態dp[i][j]代表前i個銀行，得到了j塊錢但是不被抓到最大概率。（其實設置最小概率應該也可以，但是轉移的時候不是很好寫。）

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 10000 + 5; double dp[205][MAX],P,p[205]; int n,v[205]; int main() {int t,iCase=0;cin>>t;while(t--) {scanf("%lf%d",&P,&n);int sum = 0;for(int i = 1; i<=n; i++) scanf("%d%lf",&v[i],&p[i]),sum += v[i];for(int i = 1; i<=n; i++) {for(int j = 0; j<=sum; j++) dp[i][j]=0;}dp[0][0]=1;for(int i = 1; i<=n; i++) {for(int j = 0; j<=sum; j++) {dp[i][j] = dp[i-1][j];if(j >= v[i]) dp[i][j] = max(dp[i][j],dp[i-1][j-v[i]]*(1-p[i]));}}int ans = 0 ;for(int i = 0; i<=sum; i++) {if(dp[n][i] >= 1-P) ans = i;}printf("Case %d: %d\n",++iCase,ans);}return 0 ; }

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總結

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