/*状态方程为dp[j]=max(dp[j],dp[j-1]+money[i]);
即报销j个发票所得到的最大经费,可以第j个是报销的,
也可以是第j个不报销而最大经费是由前j-1个发票加上另外第i个发票的报销数额*/
#include <stdio.h>
#include <string.h>
int main()
{
double sum,Q,asum,bsum,csum,price,ans;
double dp[35],money[35];
int n,m,num,i,j,flag;
char ch;
while(scanf("%lf %d",&Q,&n)&&n)
{
num=0;
memset(dp,0,sizeof(dp));
memset(money,0,sizeof(money));
for (i=0;i<n;i++)
{
flag=1;
sum=asum=bsum=csum=0.0;
scanf("%d",&m);
for (j=0;j<m;j++)
{
getchar();
scanf("%c:%lf",&ch,&price);
if(ch!='A'&&ch!='B'&&ch!='C'||price>600.0)
{
flag=0;
break;
}
else if(ch=='A')
asum+=price;
else if(ch=='B')
bsum+=price;
else
csum+=price;
}
sum=asum+bsum+csum;
if(flag&&sum<=1000.0&&asum<=600.0&&bsum<=600.0&&csum<=600.0)
{
money[num]=sum;
num++;
}
}
for (i=0;i<=num;i++)
{
for(j=num;j>=1;j--)
if(j==1||dp[j-1]>0&&dp[j-1]+money[i]<=Q)
dp[j]=dp[j]>(dp[j-1]+money[i])?dp[j]:(dp[j-1]+money[i]);
}
ans=0.0;
for(i=0;i<=num;i++)
if(ans<dp[i])
ans=dp[i];
printf("%.2lf\n",ans);
}
return 0;
}
hdu1844 最大报销额
最新推荐文章于 2019-03-13 16:37:59 发布