HDU 5410 CRB and His Birthday(01背包+完全背包）

CRB and His BirthdayTime Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

SubmitStatusPracticeHDU 5410

Description

Today is CRB's birthday. His mom decided to buy many presents for her lovely son.
She went to the nearest shop with   Won(currency unit).
At the shop, there are   kinds of presents.
It costs     Won to buy one present of   -th kind. (So it costs   ×     Won to buy   of them.)
But as the counter of the shop is her friend, the counter will give       candies if she buys   (   >0) presents of   -th kind.
She wants to receive maximum candies. Your task is to help her.
1 ≤   ≤ 20
1 ≤   ≤ 2000
1 ≤   ≤ 1000
0 ≤       ≤ 2000
1 ≤     ≤ 2000

 

Input

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

 

Output

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

 

Sample Input

    

     1
100 2
10 2 1
20 1 1 
    

 

Sample Output

    

     21 
    

Hint

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


解题：先跑一遍01背包，价值为a+b的，然后再按价值为a的完全背包跑。（换过来也一样）

<span style="font-size:18px;">#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int dp[2005];
int n,m,t;
int a[1005],b[1005],v[1005];
int main()
{
    scanf("%d",&t);
    while(t--){
        scanf("%d %d",&m,&n);
        for(int i=1;i<=n;i++)
            scanf("%d %d %d",&v[i],&a[i],&b[i]);
        memset(dp,0,sizeof(dp));
        int ans=0;
        for(int i=1;i<=n;i++){
            for(int j=v[i];j<=m;j++){
                dp[j]=max(dp[j],dp[j-v[i]]+a[i]);
                ans=max(ans,dp[j]);
            }
            for(int j=m;j>=v[i];j--){
                dp[j]=max(dp[j],(dp[j-v[i]]+a[i]+b[i]));
                ans=max(ans,dp[j]);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}</span>