Robberies（01背包的概率问题）

描述：

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.

输入：

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 .

输出：

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.

样例输入：

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

样列输出：

2
4

题目大意：先输入一个正整数表示测试样例，每一个测试样例包含一个正整数n（表示种数）和一个实数p（表示概率），接下去n行包含两个正整数表示价值与概率。求概率小于等于p情况下所能得到的最大价值。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct asd
{  int value;  double p;
}gw[105];
int main()
{  int t,n,allvalue,k;  double a,dp[10005];  scanf("%d",&t);  while(t--){  scanf("%lf %d",&a,&n);  a=1-a;                                                  //安全的概率 allvalue=0;  for(int i=1;i<=n;i++){  scanf("%d %lf",&gw[i].value,&gw[i].p);  gw[i].p=1-gw[i].p;  allvalue+=gw[i].value;  }  memset(dp,0,sizeof(dp));  dp[0]=1;  for(int i=1;i<=n;i++){  for(int j=allvalue;j>=gw[i].value;j--){  dp[j]=max(dp[j],dp[j-gw[i].value]*gw[i].p);         //最大情况出现的两种可能性 }  }  for(int i=allvalue;i>=0;i--){  if(dp[i]>=a){  k=i;break;  }  }  printf("%d\n",k);}  return 0;
}

Robberies（01背包的概率问题）相关推荐

HDOJ2955 Robberies(01背包，概率)
http://acm.hdu.edu.cn/showproblem.php?pid=2955 这道DP题WA了. 今天看了一下,WA在了double max(int x,int y)上,x和y都应该是 ...
HDU 1203 I NEED A OFFER! （动态规划、01背包、概率）
I NEED A OFFER! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
HDU - Robberies(01背包)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2955 Time Limit: 2000/1000 MS (Java/Others) Memory Li ...
HDOJ 2955 Robberies (01背包)
10397780 2014-03-26 00:13:51 Accepted 2955 46MS 480K 676 B C++ 泽泽 http://acm.hdu.edu.cn/showproblem. ...
暑假N天乐 —— 01背包及变形
暑假集训的第一天第二天了. 决定还是先把基础巩固一下好了,万一后面训练赛出事故更尴尬. 所以今天的主题是 -- 01背包及变形动态规划概念动态规划(Dynamic programming,DP) ...
HDU 2955 Robberies（概率DP，01背包）题解
题意:给出规定的最高被抓概率m,银行数量n,然后给出每个银行被抓概率和钱,问你不超过m最多能拿多少钱思路:一道好像能直接01背包的题,但是有些不同.按照以往的逻辑,dp[i]都是代表i代价能拿的最高 ...
01背包+概率dp Just another Robbery 抢银行
题意:有n各个银行,每个银行有v[ i ]的钱并且有x [ i ] 的被抓住啊概率,如果被抓概率<=p,那么他是安全的,求他期望的最大钱数. 输入 3 //3组 0.04 3 1 0.02 ...
hdu 2955 01背包
http://acm.hdu.edu.cn/showproblem.php?pid=2955 如果认为:1-P是背包的容量,n是物品的个数,sum是所有物品的总价值,条件就是装入背包的物品的体积和不能 ...
HDU1203_I NEED A OFFER!【01背包】
I NEED A OFFER! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...

Robberies（01背包的概率问题）

Robberies（01背包的概率问题）相关推荐

最新文章

热门文章