HDU 2955-Robberies（01背包+概率）

最新推荐文章于 2019-09-28 10:20:31 发布

I_believe_CWJ

最新推荐文章于 2019-09-28 10:20:31 发布

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分类专栏： ACM_动态规划文章标签： ACM dp 01背包

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解决一个经典的概率与收益优化问题，即如何在多个目标中选择最佳组合以最大化收益同时确保被抓概率低于可接受范围。

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Robberies

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 28110 Accepted Submission(s): 10323

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

  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

  2 4 6
 

Source

IDI Open 2009

题目大意：一个人去抢劫银行，有n个银行给他抢，抢第i个银行的利润为Mi被抓的概率为Pi，让你求在被抓的概率不超过限定概率的情况下，能抢到的钱最多是多少？

解题思路：这里牵扯到了概率问题，不知到大家还记得高中的知识吗？题目要求，不要被抓到，所以我们在抢每个银行的时候都不能被抓，所以不被抓的概率为所抢银行不被抓的概率的乘积为 p =（1-p1）*（1-p2）....，最后输出满足1-p<=maxp的最大利润就可以了。我们dp[j] = max(dp[j], dp[j-M[i]]]*(1-Pi))表示抢了j块钱不被抓的最大概率，最后我们贪心一下即可。

AC代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<set>
#include<queue>
#define N 110

using namespace std;

int T, n, m[N], maxm;
double p[N], dp[N*100], maxp;

int main()
{
    scanf("%d", &T);
    while(T --) {
        memset(dp, 0, sizeof(dp));
        scanf("%lf%d", &maxp, &n);
        maxm = 0;
        dp[0] = 1;
        for(int i = 0; i < n; i ++) {
            scanf("%d%lf", &m[i], &p[i]);
            maxm += m[i];
        }
        for(int i = 0; i < n; i ++) {
            for(int j = maxm; j >= m[i]; j --) {
                dp[j] = max(dp[j], dp[j-m[i]]*(1-p[i]));
            }
        }
        for(int i = maxm; i >= 0; i --) { //贪心输出最大
            if(dp[i] >= 1-maxp) {
                printf("%d\n", i); break;
            }
        }
    }
    return 0;
}