hdu 2955 Robberies (0 1背包)

Robberies
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8096 Accepted Submission(s): 3063

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

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gaojie

题意：给N个银行，给出每个银行里相应的钱数bank[i].money和抢这个银行被抓的概率pi。
问在不被抓的情况下(被抓的概率小于给定值p时安全)能抢得的最大钱数。

分析：
1、抢n个银行，不被抓，是每次都不被抓，用乘法原理处理每次不被抓的概率。
2、转化为0 1 背包：将所有银行的总钱数当做背包的容积v，将不被抓的概率当做cost[i]，
      在一般的0 1背包中都是加上cost[i] ，这里由于是乘法原理，所以应该是转化为乘法。
      这样，传统的01背包累加问题变成了垒乘。
3、状态转移方程：dp[j]=max(dp[j],dp[j-bank[i].money]*bank[i].run)
    dp[j]表示抢j钱时的最大不被抓概率。

代码：

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
struct node
{
    int money;
    double run;     //不被抓的概率
}bank[105];
double dp[105*105];   //数组的大小 开成105会RE

int main()
{
    double t,p;
    int T,i,n,j,maxmoney;
    scanf("%d",&T);
    while(T--)
    {
        maxmoney=0;
        scanf("%lf %d",&p,&n);
        p=1.0-p;           //最大被抓概率转化为最小不被抓概率
        for(i=1;i<=n;i++)
        {
            scanf("%d %lf",&bank[i].money,&t);
            bank[i].run=1.0-t;           //转化为不被抓的概率
            maxmoney+=bank[i].money;
        }
        memset(dp,0,sizeof(dp));       //初始化为0不是-1
        dp[0]=1.0;                   //不抢钱被抓概率为0
        for(i=1;i<=n;i++)         //抢i银行的钱
        {
            for(j=maxmoney;j>=bank[i].money;j--)
                dp[j]=max(dp[j-bank[i].money]*bank[i].run,dp[j]);
        }
        for(i=maxmoney;i>=0;i--)      //从最大钱数开始枚举
        {
            if(dp[i]>p)
            {
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}

/*
	for(i=1;i<=n;i++)
    {
		for(j=maxmoney;j>=bank[i].money;j--)
			dp[j]=max(dp[j-bank[i].money]*bank[i].run,dp[j]);
    }
	dp[j-bank[i].money]表示除了抢这个银行，之前抢的银行不被抓的概率。
	很类似组合数。
	结合第一个样例：
	这样依次算出的是dp[0](已知) dp[1] dp[3] dp[2] dp[6] dp[5] dp[4]

*/

补充：软件开发 , C++ ,