HDOJ Robberies（好题背包）

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16433    Accepted Submission(s): 6033

Problem Description

The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usually gets caught in the end, often because they become too greedy. He has decided to work in the lucrative business of bank robbery only for a short while, before retiring to a comfortable job at a university.


For a few months now, Roy has been assessing the security of various banks and the amount of cash they hold. He wants to make a calculated risk, and grab as much money as possible.


His mother, Ola, has decided upon a tolerable probability of getting caught. She feels that he is safe enough if the banks he robs together give a probability less than this.

 

Input

The first line of input gives T, the number of cases. For each scenario, the first line of input gives a floating point number P, the probability Roy needs to be below, and an integer N, the number of banks he has plans for. Then follow N lines, where line j gives an integer Mj and a floating point number Pj .
Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .

 

Output

For each test case, output a line with the maximum number of millions he can expect to get while the probability of getting caught is less than the limit set.

Notes and Constraints
0 < T <= 100
0.0 <= P <= 1.0
0 < N <= 100
0 < Mj <= 100
0.0 <= Pj <= 1.0
A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

 

Sample Input

30.04 31 0.022 0.033 0.050.06 32 0.032 0.033 0.050.10 31 0.032 0.023 0.05

 

Sample Output

246因为逃跑率被捕率都是小数，所以普通的背包是不行了，刚开始也想过吧概率x100什么的，那样就变成了01背包，但是不行，然后开始换思路，把银行的钱看成是价值，把几率看成是重量，然后方程就是：dp[j]=max(dp[j],dp[j-a[i].v]+a[i].p)输出一看，诶，和样例也一样，嗯，提交，wrong！！！！回来继续看，在纸上列了一下，发现问题了，明明说了是概率为什么要加呢，并不是那么简单的，所以1-a[i].p就是逃跑的概率咯，应该是乘的吧，所以方程就是：dp[j]=max(dp[j],dp[j-a[i].v]*(1-a[i].p))然后最后只要逃跑率大于被捕率就行了。ac代码：
#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#define MAXN 10010#define INF 0xfffffff#define max(a,b) a>b?a:b#define min(a,b) a>b?b?ausing namespace std;struct s{double p;int v;}a[MAXN];int main(){int t,n;int i,j,sum;double m;scanf("%d",&t);while(t--){scanf("%lf%d",&m,&n);sum=0;for(i=0;i<n;i++){scanf("%d%lf",&a[i].v,&a[i].p);sum+=a[i].v;}double dp[MAXN];memset(dp,0,sizeof(dp));dp[0]=1;for(i=0;i<n;i++){for(j=sum;j>=a[i].v;j--){//dp[j]=max(dp[j],dp[j-a[i].v]+a[i].p);dp[j]=max(dp[j],dp[j-a[i].v]*(1-a[i].p));}}for(i=sum;i>=0;i--){if(dp[i]>(1-m))//逃跑率大于被捕率 {printf("%d\n",i);break;}}}return 0;}