2015暑假多校联合---CRB and His Birthday（01背包）

题目链接

http://acm.split.hdu.edu.cn/showproblem.php?pid=5410

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

100 2

10 2 1

20 1 1

Sample Output

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

Recommend

wange2014

题意：输入M，N，分别表示总的钱数和物品种数，接下来输入N行，每行3个数，单价、买一件送的糖数、买一次送的糖数求最多能得到多少糖？

思路：01背包，dp[i]表示i钱下能得到最多的糖数，vis[i][j]表示i钱下得到最多糖时，是否买j物品，状态转移方程dp[i]=dp[i-kind[j][0]]+kind[j][1]+(vis[i-kind[j][0]][j]==0) * kind[j][2]；

代码如下：

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <cstring>

#include <cmath>

#define eps 1e-8

#define maxn 105

#define inf 0x3f3f3f3f3f3f3f3f

#define IN freopen("in.txt","r",stdin);

using namespace std;

int dp[];

int vis[][];

int kind[][];

int main()

{

    int T;

    int M,N;

    cin>>T;

    while(T--)

    {

        scanf("%d%d",&M,&N);

        for(int i=;i<N;i++)

            scanf("%d%d%d",&kind[i][],&kind[i][],&kind[i][]);

        memset(dp,,sizeof(dp));

        memset(vis,,sizeof(vis));

        for(int i=;i<=M;i++)

        {

            int flag=-;

            for(int j=;j<N;j++)

            {

                if(i<kind[j][]) continue;

                int s=dp[i-kind[j][]]+kind[j][];

                if(!vis[i-kind[j][]][j]) s+=kind[j][];

                if(dp[i]<s)

                {

                    dp[i]=s;

                    flag=j;

                }

            }

            if(flag>=)

            {

                for(int j=;j<N;j++)

                {

                    vis[i][j]=vis[i-kind[flag][]][j];

                }

                vis[i][flag]++;

            }

        }

        int tmp=;

        for(int i=;i<=M;i++)

           tmp=max(tmp,dp[i]);

        printf("%d\n",tmp);

    }

    return ;

}

巴特西

2015暑假多校联合---CRB and His Birthday（01背包）

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