HDU 5410 CRB and His Birthday（完全背包变形）

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

 

 

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

 

 

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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=1;i<=n;i++)
        scanf("%d%d%d",&w[i],&a[i],&b[i]);

        memset(f,0,sizeof(f));

        for(int i=1;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 0;
}

 

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=1;i<=n;i++)
        scanf("%d%d%d",&w[i],&a[i],&b[i]);

        memset(f,0,sizeof(f));
        memset(g,0,sizeof(g));

        for(int i=1;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=0;j<=m;j++)
            {
                g[j]=f[j];
            }
        }
        cout<<f[m]<<endl;
    }
    return 0;
}

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 = 2005;
int dp[N][N][2];

int main ()
{
    int t;cin>>t;
    while ( t-- )
    {
        int m, n;
        scanf("%d%d", &m, &n);
        memset( dp, 0, sizeof(dp) );
        for ( int i = 1; i <= n; i++ )
        {
            int w, a, b;
            scanf("%d%d%d", &w, &a, &b);
            for ( int j = 0; j <= m; j++ )
            {
                dp[i][j][0] = max( dp[i-1][j][0], dp[i-1][j][1] );
            }
            for ( int j = w; j <= m; j++ )
            {
                dp[i][j][1] = max( dp[i][j - w][1] + a, dp[i][j - w][0] + a + b );
            }
        }
        printf("%d\n", max( dp[n][m][0], dp[n][m][1] ));
    }
    return 0;
}

 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 = 0; j <= m; j++ )
    {
        dp[i][j][0] = max( dp[i-1][j][0], dp[i-1][j][1] );
    }
    for ( int j = w; j <= m; j++ )
    {
        dp[i][j][1] = max( dp[i][j - w][1] + a, dp[i][j - w][0] + a + b );
    }