hdu 5410 CRB and His Birthday 2015多校联合训练赛#10 dp 动态规划

CRB and His Birthday

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 132    Accepted Submission(s): 72

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 integer  T , 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

   
   
   
   

    
    
    
    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. 
    
    
    
    
     
   
   
   
   

 

Author

KUT（DPRK）

 

Source

2015 Multi-University Training Contest 10

背包问题。

特判每个状态是否可以由dp[i-cost]转移即可

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
#define maxn 2007
int dp[maxn];
int res[maxn];
int check[maxn];
int main(){
    int t, m,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&m,&n);
        memset(dp,0,sizeof(dp));
        int cost,a,b;
        for(int i = 0;i < n; i++){
            scanf("%d%d%d",&cost,&a,&b);
            memset(check,0,sizeof(check));
            for(int i = 0;i <= m; i++)
                res[i] = dp[i];
            for(int i = cost; i <= m; i++){
                if(res[i-cost] + a + b>= dp[i])
                    check[i] = 1;
                res[i] = max(res[i],res[i-cost]+a);
            }
            for(int i = m;i >= 0; i--)
                if(check[i] == 1)
                    res[i] = max(res[i],res[i-cost]+a+b);
            for(int i = 0;i <= m; i++)
                dp[i] = max(dp[i],res[i]);
        }
        int ans = 0;
        for(int i = 0;i <= m; i++)
            ans = max(ans,dp[i]);
        printf("%d\n",ans);
    }
    return 0;
}