HDU 2955 Robberies
2015-11-28 01:01
253 查看
Robberies
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 17647 Accepted Submission(s): 6516
[align=left]Problem Description[/align]
The aspiring(有抱负的) Roy the Robber has seen a lot of American movies, and knows that the bad
guys(球员) usually gets caught in the end, often because they become too greedy. He has decided to work in the
lucrative(有利可图的) 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(评定) the security of various banks and the
amount(数量) of cash they hold. He wants to make a
calculated(计算出的)
risk(风险), and grab as much money as possible.
His mother, Ola, has decided upon a tolerable(可以的)
probability(可能性) of getting caught. She feels that he is safe enough if the banks he robs together give a probability less than this.
[align=left]Input[/align]
The first line of
input(投入) gives T, the number of cases. For each
scenario(方案), the first line of input gives a floating point number P, the
probability(可能性) Roy needs to be below, and an
integer(整数) 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 .
[align=left]Output[/align]
For each test case,
output(输出) 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(破产者) if it is robbed, and you may
assume(承担) that all probabilities are independent as the police have very low funds.
[align=left]Sample Input[/align]
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
[align=left]Sample Output[/align]
2
4
6
题目意思是给定给定一个概率,是被抓的概率,给定几个银行,每个银行对应一个钱数和被抓的概率,如何在有效的被抓概率内偷更多的钱?
显然是个0-1背包的问题,一开始把这个当成了一个概率简单相加的问题,看了讨论区发现傻了。。其实概率应该是相乘的。设个一维DP,代表对应money被抓的概率,最后反向寻找第一个概率大于1 - 被抓的概率的就好。
代码如下:
/*************************************************************************
> File Name: 0-1_bags.cpp
> Author: Zhanghaoran
> Mail: chilumanxi@xiyoulinux.org
> Created Time: Fri 27 Nov 2015 11:21:43 PM CST
************************************************************************/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;
float dp[10010];
int main(void){
int T, n;
float p;
int mon[10010];
float pro[10010];
int all = 0;
cin >> T;
while(T --){
all = 0;
scanf("%f%d",&p,&n);
memset(dp, 0, sizeof(dp));
for(int i = 1; i <= n; i ++){
scanf("%d%f",&mon[i],&pro[i]);
all += mon[i];
}
dp[0] = 1;
for(int i = 1; i <= n; i ++){
for(int j = all; j >= mon[i]; j --){
dp[j] = max(dp[j], dp[j - mon[i]] * (1 - pro[i]));
}
}
for(int i = all; i >= 0; i --){
if(dp[i] >= (1 - p)){
printf("%d\n", i);
break;
}
}
}
return 0;
}
相关文章推荐
- 动易2006序列号破解算法公布
- Ruby实现的矩阵连乘算法
- C#插入法排序算法实例分析
- 超大数据量存储常用数据库分表分库算法总结
- C#数据结构与算法揭秘二
- C#冒泡法排序算法实例分析
- 算法练习之从String.indexOf的模拟实现开始
- C#算法之关于大牛生小牛的问题
- C#实现的算24点游戏算法实例分析
- c语言实现的带通配符匹配算法
- 浅析STL中的常用算法
- 算法之排列算法与组合算法详解
- C++实现一维向量旋转算法
- Ruby实现的合并排序算法
- C#折半插入排序算法实现方法
- 基于C++实现的各种内部排序算法汇总
- C++线性时间的排序算法分析
- C++实现汉诺塔算法经典实例
- PHP实现克鲁斯卡尔算法实例解析
- C#获取关键字附近文字算法实例