hdu 5410 01+完全背包

CRB and His Birthday

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1257    Accepted Submission(s): 615

Problem Description

Today is CRB's birthday. His mom decided to buy many presents for her lovely son.
She went to the nearest shop with M Won(currency unit).
At the shop, there are N kinds of presents.
It costs Wi Won to buy one present of i-th kind. (So it costs k × Wi Won to buy k of them.)
But as the counter of the shop is her friend, the counter will give Ai × x + Bi candies if she buys x(x>0) presents of i-th kind.
She wants to receive maximum candies. Your task is to help her.
1 ≤ T ≤ 20
1 ≤ M ≤ 2000
1 ≤ N ≤ 1000
0 ≤ Ai, Bi ≤ 2000
1 ≤ Wi ≤ 2000

 

Input

There are multiple test cases. The first line of input contains an integerT, indicating the number of test cases. For each test case:
The first line contains two integers M and N.
Then N lines follow, i-th line contains three space separated integers Wi,Ai and Bi.

 

Output

For each test case, output the maximum candies she can gain.

 

Sample Input

1100 210 2 120 1 1

 

Sample Output

21
Hint
CRB's mom buys 10 presents of first kind, and receives 2 × 10 + 1 = 21 candies.
 

 

Author

KUT（DPRK）

 

题意：有n种商品，每种商品中有a个糖果，如果买这种商品就送多b个糖果，只有第一次买的时候才送。现在有m元，最多能买多少糖果？

思路：第一次买一种商品时有送糖果，对这一次进行一次01背包，也就是只能买一次。然后对这种商品来一次完全背包，此时不送糖果，也可以多买。

ac代码：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<cctype>#include<stdio.h>#define min(a,b)(a<b?a:b)#define max(a,b)(a>b?a:b)#define INF 0x3f3f3f3ftypedef long long ll;using namespace std;int a[1100],w[1100],b[1100];int dp[2100];int main(){    int T,m,n,i,j;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&m,&n);        for(i=1;i<=n;i++)        {            scanf("%d%d%d",&w[i],&a[i],&b[i]);        }        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++)        {            for(j=m;j>=w[i];j--)            {                dp[j]=max(dp[j],dp[j-w[i]]+a[i]+b[i]);            }            for(j=w[i];j<=m;j++)            {                dp[j]=max(dp[j],dp[j-w[i]]+a[i]);            }        }        ll sum=-1;        for(i=m;i>0;i--)            sum=max(sum,dp[i]);        printf("%I64d\n",sum);    }    return 0;}