LightOJ 1079 - Just another Robbery (01背包)

1079 - Just another Robbery

PDF (English)

Statistics

Forum

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.

题意：XX想抢劫银行，当危险率低于P的时候才能行动，现在给出每家银行的金钱mi和危险率pi，求最多能获得多少金钱。

题解：危险率是P，那么安全率就是1-P，那么XX抢劫的所有银行的安全率之积就不能小于1-P，这样就变成了一个01背包的裸题。我真是笨啊，背包看了两次，做了十几道题了，比赛的时候居然没想到用背包解(；′⌒`)

代码如下：

#include<cstdio>
#include<cstring>
double dp[10010];
int m[110];
double p[110];

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

int main()
{
int t,k=1,i,j,n,sum;
double pp;
scanf("%d",&t);
while(t--)
{
scanf("%lf%d",&pp,&n);
pp=1.0-pp;
sum=0;
for(i=0;i<n;++i)
{
scanf("%d%lf",&m[i],&p[i]);
sum+=m[i];
p[i]=1.0-p[i];
}
memset(dp,0,sizeof(dp));
dp[0]=1;
for(i=0;i<n;++i)
{
for(j=sum;j>=m[i];--j)
dp[j]=max(dp[j],dp[j-m[i]]*p[i]);
}
for(i=sum;i>=0;--i)
{
if(dp[i]>=pp)
break;
}
printf("Case %d: %d\n",k++,i);
}
return 0;
}