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原题http://acm.hdu.edu.cn/showproblem.php?pid=2955

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11820    Accepted Submission(s): 4398

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.

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

2
4
6

//题目大意，给你一个概率，然后n组数据，每组数据的整数代表钱，小数代表被抓的概率
//问，在不被抓的概率的前提下，最多能拿多少钱。
//思路，开个dp数组，代表钱在n的时候被抓的不被抓的概率，
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <set>
#include <map>
using namespace std;
#define N 100 + 10
double dp[10009];
int sum[N];
double p[N];

double max(double a,double b)
{
    return a>b?a:b;
}

int main()
{
    int T,n,i;
    double v;

    while(~scanf("%d",&T))
    {
        while(T--)
        {
            memset(dp,0,sizeof(dp));
            dp[0] = 1;
            memset(sum,0,sizeof(sum));
            memset(p,0,sizeof(p));
            scanf("%lf%d",&v,&n);
            int mark = 0;
            for(i=1; i<=n; i++)
            {
                scanf("%d%lf",&sum[i],&p[i]);
                mark+=sum[i];
                p[i] = 1-p[i];
            }
            int j;
            for(i=0;i<=mark;i++){
                dp[i] = 0;
            }
            dp[0] =1;//当抢了0元时，肯定是安全的。所以初始化为1
            for(i=1; i<=n; i++)
            {
                for(j=mark; j>=sum[i]; j--)
                {
                    dp[j] = max(dp[j],dp[j-sum[i]]*(p[i]));
                }
            }
            v = 1-v;
            for(i=mark; i>=0; i--)
            {
                if(dp[i] >= v){
                    printf("%d\n",i);
                    break;

                }

            }
        }
    }

    return 0;
}

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01背包，概率

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原文地址：http://blog.csdn.net/zcr_7/article/details/38488681