HDOJ 2955 Robberies (01背包)
10397780 | 2014-03-26 00:13:51 | Accepted | 2955 | 46MS | 480K | 676 B | C++ | 泽泽 |
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): 9836 Accepted Submission(s): 3673
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.
Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .
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.
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
4
6
#include<stdio.h>
#include<string.h>
int main()
{
int v[];
double f[],w[];
int i,j,n,m;
int T;
scanf("%d",&T);
while(T--)
{
double rate;
int sum=;
scanf("%lf %d",&rate,&n);//rate被抓的概率,1-rate逃脱的概率
for(i=;i<n;i++)
{
scanf("%d %lf",&v[i],&w[i]);
sum+=v[i];
}
memset(f,,sizeof(f));
f[]=;//没偷逃脱的概率为1
for(i=;i<n;i++)
{
for(j=sum;j>=v[i];j--)
{
f[j]=f[j]>f[j-v[i]]*(-w[i])?f[j]:f[j-v[i]]*(-w[i]);//选择最大逃脱概率
}
}
for(i=sum;i>=;i--)
{
if(f[i]>=-rate)
{
printf("%d\n",i);
break;
}
}
}
return ;
}
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