HDU 2955 Robberies 背包概率DP

最新推荐文章于 2019-09-28 10:20:31 发布

转载最新推荐文章于 2019-09-28 10:20:31 发布 · 91 阅读

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原文链接：http://www.cnblogs.com/13224ACMer/p/4983177.html

本文探讨了一道经典的01背包问题，通过分析银行抢劫案例，讲解如何在限制被抓概率的前提下，计算能抢夺的最大金额。文章详细解释了动态规划算法的应用，并提供了完整的代码实现。

A - Robberies

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2955

Appoint description:

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

Sample Output

2

4

6

   这是一道01背包的问题，但是有一个地方我没有想到，所以花了很多时间。 
   　　题目给了每个银行的钱和被抓的概率，由于要抢尽量多的钱，所以要保证尽量不被抓，而抢多个银行之后不被抓的概率是抢每一个银行不被抓的概率之 积，我竟然把这一点给忘了！导致我走了许多弯路，思路不能太死啊！dp[]表示抢到下标所对应的钱时，此时不被抓的概率，题目给出了最终不能高于被抓概率 P，不被抓的概率就不能低于（1-P），所以最后只需要逆序遍历dp,找到第一个大于等于1-P的dp[i]，就能够保证i最大，即抢到的钱最多。
 
       /* ***********************************************
   Author        :Mubaixu
   File Name     :HDU2955.cpp
   ************************************************ */

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
#include<algorithm>
using  namespace std;
const int maxm=10005;
const int maxn=105;
double dp[maxm];
int value[maxn];
double tp[maxn];
int main(){
     int t;
    int n;
    double p;
    scanf("%d",&t);
    while(t--){
         scanf("%lf%d",&p,&n);
     
        int sum=0;
        for(int i=1;i<=n;i++){
             scanf("%d%lf",&value[i],&tp[i]);
            tp[i]=1-tp[i];
            sum+=value[i];
        }
         memset(dp,0,sizeof(dp));
        dp[0]=1;

        for(int i=1;i<=n;i++){
           for(int j=sum;j>=value[i];j--){
               dp[j]=max(dp[j],dp[j-value[i]]*tp[i]);
           }
        }
          for(int i=sum;i>=0;i--){
             if(dp[i]>(1-p)){
                 printf("%d\n",i);
                 break;
             }

          }

    }
    return 0;
} 
   
 

转载于:https://www.cnblogs.com/13224ACMer/p/4983177.html