\n int asp rabl wan limit line esc std

Robberies

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31769 Accepted Submission(s): 11527

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

題目大意：

Roy是一個盜賊。現在有n個銀行成了他的下手目標。對第i個銀行下手，他會獲得mi元錢，同時有pi的概率被抓。問在被抓概率不超過p的情況下，他最多能得到多少錢。

01背包好題。

這題仔細想想還是很有趣的。獲得的錢其實是背包容量。獲得這些錢安全的概率才是dp存儲的東西。要好好想想。

此外，對double類型變量輸入輸出是 printf&%f scanf&%lf，註意。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<stack>

using namespace std;

const int maxn=100;

int money[maxn+5];
double safe[maxn+5];//不被抓的概率作為重量
double dp[maxn*maxn+5];//獲得i錢，不被抓的概率

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        double p;int n;
        scanf("%lf%d",&p,&n);
        p=1-p;
        int sum=0;//最多獲得多少錢
        for(int i=1;i<=n;i++)
        {
            scanf("%d%lf",money+i,safe+i);
            safe[i]=1-safe[i];
            sum+=money[i];
        }

        for(int i=0;i<=sum;i++)
            dp[i]=0;
        dp[0]=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=sum;j>=money[i];j--)
            {
                dp[j]=max(dp[j],dp[j-money[i]]*safe[i]);
            }
        }

        for(int i=sum;i>=0;i--)
            if(dp[i]>p)
            {
                printf("%d\n",i);
                break;
            }
    }
    return 0;
}

View Code

hdu 2955 Robberies (01背包好題)