MM enjoyed cookies very much. On Saint Valentine‘s Day, when she stepped into a big cookie store again, she wouldn‘t leave unless DD spent all his money in pocket!
There are N kinds of cookies, labeled from 1 to N, and all can be bought without any restriction by the store. But actually, for some kinds of cookies, MM wanted to buy one piece at most, and for some kinds of cookies, MM wanted to buy Ki pieces at most, and for some other kinds of cookies, there didn‘t exist an upper bound that MM wanted to buy.
There is another requirement from MM: there are some groups of cookies, MM considered their tastes similar, so she wanted to buy at most one kind of cookies in each group. A kind of cookie wouldn‘t appear in more than one group.
For the ith kind of cookies, MM has an "enjoyable value" Ei, if DD bought Ai pieces of this kind for her, and Ai didn‘t exceed her upper bound, MM get EiAi of enjoyable value. After buying cookies, MM‘s total enjoyable value will be the sum of EiAi.
But actually, poor DD had only D dollars, and the price for the ith kind of cookies is Pi dollars per piece. DD must spend all his D dollars to buy cookies, to meet requirements about amount and taste from MM, and to make MM‘s enjoyable value as high as possible. What‘s more, as you know, a legal plan‘s enjoyable value must be non-negative.
Input
There are multiple test cases. Each test case consists of three parts.
The first part is one line with two integers N and D.
The second part has N lines, line i consists of three integers Ki, Ei and Pi. If Ki equals to 0, it means for ith kind of cookies, there didn‘t exist an upper bound that MM wanted to buy, otherwise Ki is the upper bound for ith kind of cookies.
The third part describes the groups. A non-negative integer G represents the number of groups, and then G lines, each line consists of some integers represents labels of kinds of cookies in this group.
One blank line between test cases.
Output
If the proper and optimal plan exists, output the maximal total enjoyable value ΣEiAi, otherwise output "i‘m sorry...".
Output one line per text case.
Data Restriction
1 <= N <= 1024, 0 <= D <= 1024.0 <= Ki <= 1024, -1024 <= Ei <= 1024, 0 < Pi <= D.
0 <= G <= 8.
All numbers referred are integers.
Number of test cases is no more than 80.
Sample Input
2 1024 0 1 3 0 0 1 0 10 1023 1 1 1 1 1 2 1 1 4 1 1 8 1 1 16 1 1 32 1 1 64 1 1 128 3 -1 256 1 1 512 1 9 10 10 1023 1 1 1 1 1 2 1 1 4 1 1 8 1 1 16 1 1 32 1 1 64 1 1 128 1 1 256 1 1 512 1 9 10
Sample Output
341 5 i‘m sorry...
题意：
给你一些物品和一些钱，每种物品有价值、个数、价格，然后物品可能会分组，每组中的物品至多拿一件，问将这些钱花光，得到的最大价值。
思路：
典型的分组混合背包问题。第一次做，参考了别人的代码，得先懂混合背包和分组背包后做。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 2005
#define MAXN 16005
#define mod 1000000000
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;
int n,m,ans,cnt,tot,flag;
int f[maxn],dp[maxn];
int k[maxn],e[maxn],p[maxn];
int gro[maxn],w[10][maxn]; // w[i][j]-第i组花费j的钱能得到的最大价值
char s[maxn];
void ZeroOnePack(int cost,int val,int tmp[])
{
for(int i=m;i>=cost;i--)
{
tmp[i]=max(tmp[i],tmp[i-cost]+val);
}
}
void CompletePack(int cost,int val,int tmp[])
{
for(int i=cost;i<=m;i++)
{
tmp[i]=max(tmp[i],tmp[i-cost]+val);
}
}
void MultiplePack(int cost,int val,int num,int tmp[])
{
int i,j,t=0;
for(j=0; ;j++)
{
t+=(1<<j);
if(t>=num) break ;
ZeroOnePack((1<<j)*cost,(1<<j)*val,tmp);
}
t-=(1<<j);
t=num-t;
ZeroOnePack(t*cost,t*val,tmp);
}
bool solve()
{
int i,j,t,g;
memset(dp,-INF,sizeof(dp));
memset(w,-INF,sizeof(w));
int OO=dp[0];
dp[0]=0;
for(i=1;i<=n;i++)
{
if(gro[i]) // 如果没分组 则直接对dp背包 分组则对f背包后存入w
{
memset(f,-INF,sizeof(f));f[0]=0;
}
if(!k[i]||k[i]*p[i]>=m)
{
CompletePack(p[i],e[i],gro[i]?f:dp);
}
else
{
MultiplePack(p[i],e[i],k[i],gro[i]?f:dp);
}
if(gro[i])
{
for(j=0;j<=m;j++) w[gro[i]][j]=max(w[gro[i]][j],f[j]);
}
}
for(int g=1;g<=cnt;g++) // 将w[i][j]看做物品 进行分组01背包
{
for(i=m;i>=1;i--)
{
for(j=1;j<=i;j++)
{
if(dp[i-j]>OO&&w[g][j]>OO) dp[i]=max(dp[i],dp[i-j]+w[g][j]);
}
}
}
ans=dp[m];
if(ans<0) return false ;
return true ;
}
int main()
{
int i,j,t;
while(~scanf("%d%d",&n,&m))
{
for(i=1; i<=n; i++)
{
scanf("%d%d%d",&k[i],&e[i],&p[i]);
}
memset(gro,0,sizeof(gro));
scanf("%d",&cnt);
gets(s);
for(i=1; i<=cnt; i++)
{
gets(s);
for(j=0; s[j]!=‘\0‘;)
{
if(s[j]>=‘0‘&&s[j]<=‘9‘)
{
t=0;
while((s[j]>=‘0‘&&s[j]<=‘9‘))
{
t=t*10+s[j]-‘0‘;
j++;
}
gro[t]=i;
}
else j++;
}
}
if(solve()) printf("%d\n",ans);
else printf("i‘m sorry...\n");
}
return 0;
}