这道题目以前也看过,但是没学多重背包,就没做,今天做一下,思路要清晰,一定要搞清楚,谁是背包容量,这里背包容量就是所带的钱,单价是cost,粮食的重量是value,
代码如下:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=20005;
int value[maxn],cost[maxn],f[maxn];
int acount[maxn];
int t,n,m;
void zeropack(int cost1,int value1)
{
for(int i=n; i>=cost1; i--)
{
f[i]=max(f[i],f[i-cost1]+value1);
}
}
void completpack(int cost1, int value1)
{
for(int i=cost1; i<=n; i++)
{
f[i]=max(f[i],f[i-cost1]+value1);
}
}
void multiplepack(int cost1,int value1,int acount1)
{
if(cost1*acount1>=n)
{
completpack(cost1,value1);
return ;
}
else
{
int k=1;
while(k<acount1)
{
zeropack(k*cost1,k*value1);
acount1-=k;
k*=2;
}
zeropack(acount1*cost1,acount1*value1);
}
}
int main()
{
int i;
scanf("%d",&t);
while(t--)
{
memset(cost,0,sizeof(cost));
memset(value,0,sizeof(value));
memset(acount,0,sizeof(acount));
memset(f,0,sizeof(f));
scanf("%d%d",&n,&m);
for(i=1; i<=m; i++)
{
scanf("%d%d%d",&cost[i],&value[i],&acount[i]);
}
for(i=1; i<=m; i++)
{
multiplepack(cost[i],value[i],acount[i]);
}
printf("%d\n",f[n]);
}
return 0;
}