hdu 5410 CRB and His Birthday（0-1背包+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): 1605    Accepted Submission(s): 755

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）

解题思路：一个0-1背包和完全背包的融合，每种物品只有买第一种的时候会不同，之后完全一样，所以只需要在完全背包前面再加一个完全背包就好了~（^_^）

注意两种不同循环的方向~

#include <iostream>#include <cstring>#include <cstdio>using namespace std;#define maxn 1005int t,n,m;int w[maxn],a[maxn],b[maxn];int dp[2005];int main(){    int i,j;    cin>>t;    while(t--)    {        cin>>m>>n;        for(i=0; i<n; i++)            cin>>w[i]>>a[i]>>b[i];        memset(dp,0,sizeof dp);        for(i=0; i<n; i++)        {            for(j=m; j>=w[i]; j--)//0-1            {                dp[j] = max(dp[j],dp[j-w[i]]+a[i]+b[i]);            }            for(j=w[i]; j<=m; j++)//完全背包            {                dp[j] = max(dp[j],dp[j-w[i]]+a[i]);            }        }        cout << dp[m] << endl;    }    //cout << "Hello world!" << endl;    return 0;}