CRB and His Birthday
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 430 Accepted Submission(s): 237
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
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.
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
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
Recommend
大致三种方法：
1. 01背包+完全背包
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。
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 2005
#define M 12 int n,m;
int f[N],w[N],a[N],b[N]; int main()
{
int T;cin>>T;
while(T--)
{
scanf("%d%d",&m,&n);
for(int i=;i<=n;i++)
scanf("%d%d%d",&w[i],&a[i],&b[i]); memset(f,,sizeof(f)); for(int i=;i<=n;i++)
{
for(int j=m;j>=w[i];j--)
{
f[j]=max(f[j],f[j-w[i]]+a[i]+b[i]);
}
for(int j=w[i];j<=m;j++)
{
f[j]=max(f[j],f[j-w[i]]+a[i]);
} }
cout<<f[m]<<endl;
}
return ;
}
2. 完全背包的思路。每次有三种选择方案，1不选第i件物品，2选一个第2件物品，3选2个以上的第二件物品。
用一个辅助数组g记录上一行（上一个i）的最大糖果数的值，因为完全背包空间是从小到大的，所以同一个i后面的空间会用到前面的的值，这也就是完全背包的内涵，但是用了辅助数组g，记录上一行的状态，就保证使用a+b不会用到前面的值。
转移方程：f[j]=max3(f[j], f[j-w[i]]+a[i], g[j-w[i]]+a[i]+b[i]);
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 2005
#define M 12 int n,m;
int w[N],a[N],b[N];
int f[N];//f[j]表示j元钱可以得到的最大糖果数
int g[N];//g[j]表示上一个i,j元可以得到的最大糖果数 int max3(int a,int b,int c)
{
return max(a,max(b,c));
} int main()
{
int T;cin>>T;
while(T--)
{
scanf("%d%d",&m,&n);
for(int i=;i<=n;i++)
scanf("%d%d%d",&w[i],&a[i],&b[i]); memset(f,,sizeof(f));
memset(g,,sizeof(g)); for(int i=;i<=n;i++)
{
for(int j=w[i];j<=m;j++)
{
f[j]=max3(f[j], f[j-w[i]]+a[i], g[j-w[i]]+a[i]+b[i]);
}
for(int j=;j<=m;j++)
{
g[j]=f[j];
}
}
cout<<f[m]<<endl;
}
return ;
}
3. 还是完全背包的思路，用dp[i][j][0]表示不取第i件物品，花费j得到的最大收益，dp[i][j][1]表示取第i件物品，花费j得到的最大收益，最终的结果就是max(dp[i][j][0],dp[i][j][1])。每次对于物品i，先得出不取它能得到的最大收益，然后再求取它能得到的最大收益。
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std; const int N = ;
int dp[N][N][]; int main ()
{
int t;cin>>t;
while ( t-- )
{
int m, n;
scanf("%d%d", &m, &n);
memset( dp, , sizeof(dp) );
for ( int i = ; i <= n; i++ )
{
int w, a, b;
scanf("%d%d%d", &w, &a, &b);
for ( int j = ; j <= m; j++ )
{
dp[i][j][] = max( dp[i-][j][], dp[i-][j][] );
}
for ( int j = w; j <= m; j++ )
{
dp[i][j][] = max( dp[i][j - w][] + a, dp[i][j - w][] + a + b );
}
}
printf("%d\n", max( dp[n][m][], dp[n][m][] ));
}
return ;
}
第一位空间可以压缩，把声明改成 int dp[N][2];最后结果改成max(dp[m][0],dp[m][1]).
中间关键代码换成
for ( int j = ; j <= m; j++ )
{
dp[i][j][] = max( dp[i-][j][], dp[i-][j][] );
}
for ( int j = w; j <= m; j++ )
{
dp[i][j][] = max( dp[i][j - w][] + a, dp[i][j - w][] + a + b );
}