LightOJ 题目1079 Just another Robbery（01背包）

1079 - Just another Robbery

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Time Limit: 4 second(s)

Memory Limit: 32 MB

As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He wants to make a calculated risk, and grab as much money as possible. But his friends - Hermione and Ron have
decided upon a tolerable probability P of getting caught. They feel that he is safe enough if the banks he robs together give a probability less than P.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a real number P, the probability Harry needs to be below, and an integer N (0 < N ≤ 100), the number of banks he has plans for. Then follow N lines, where line j gives
an integer Mj(0 < Mj ≤ 100) and a real number Pj . Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj.
A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

Output

For each case, print the case number and the maximum number of millions he can expect to get while the probability of getting caught is less than P.

Sample Input	Output for 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	Case 1: 2 Case 2: 4 Case 3: 6

Note

For the first case, if he wants to rob bank 1 and 2, then the probability of getting caught is 0.02 + (1 - 0.02) * .03 = 0.0494 which is greater than the given probability (0.04). That's why he has only option, just to rob
rank 2.

题目大意：

一个人抢钱，每个物品有一个危险概率（double），一个价值，问低于p危险概率下，获得的最大收益

ac代码

#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<math.h>
#include<limits.h>
#define N 100010
#define mod 1000000
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
int w[110];
double v[110];
double dp[1000100];
int main()
{
double p;
int n,t,cas=0;
scanf("%d",&t);
while(t--)
{
int i,j;
scanf("%lf%d",&p,&n);
int sum=0;
for(i=1;i<=n;i++)
{
scanf("%d%lf",&w[i],&v[i]);
v[i] = 1-v[i];
sum+=w[i];
}
memset(dp,0,sizeof(dp));
dp[0]=1;
for(i=1;i<=n;i++)
{
for(j=sum;j>=w[i];j--)
dp[j]=max(dp[j],dp[j-w[i]]*v[i]);
}
for(i=sum;i>=0;i--)
{
if(1-dp[i]<=p)
break;
}
printf("Case %d: %d\n",++cas,i);
}
}

PROBLEM SETTER: JANE ALAM JAN