2015 summer vacation multi-School Cooperation --- CRB and His Birthday (01 backpack), 2015 --- crb

2015 summer vacation multi-School Cooperation --- CRB and His Birthday (01 backpack), 2015 --- crb

Question Link

Http://acm.split.hdu.edu.cn/showproblem.php? Pid = 1, 5410

 

Problem DescriptionToday is CRB's birthday. His mom decided to buy too 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 ≤ MB ≤ 2000
1 ≤ N ≤1000
0 = Ai, Bi = 2000
1 ≤ Wi ≤2000
  InputThere 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.
  OutputFor each test case, output the maximum candies she can gain. Sample Input1100 210 2 120 1 1 Sample Output21 HintCRB's mom buys 10 presents of first kind, and between ES 2 × 10 + 1 = 21 candies. AuthorKUT (DPRK) Source2015 Multi-University Training Contest 10 RecommendIn wange2014, enter M and N, respectively, indicating the total amount of money and the number of items. Then, enter N rows, with 3 numbers in each row, the unit price, the quantity of sugar to be bought, and the quantity of sugar to be bought at a time. How much sugar can be obtained at most? Idea: 01 backpack, dp [I] indicates the maximum amount of sugar that can be obtained under I, vis [I] [j] indicates the maximum amount of sugar that can be obtained under I, whether to buy j items, state transition equation dp [I] = dp [I-kind [j] [0] + kind [j] [1] + (vis [I-kind [j] [0]] [j] = 0) * kind [j] [2]; the Code is as follows:

#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[2005];int vis[2005][1005];int kind[1005][3];int main(){    int T;    int M,N;    cin>>T;    while(T--)    {        scanf("%d%d",&M,&N);        for(int i=0;i<N;i++)            scanf("%d%d%d",&kind[i][0],&kind[i][1],&kind[i][2]);        memset(dp,0,sizeof(dp));        memset(vis,0,sizeof(vis));        for(int i=1;i<=M;i++)        {            int flag=-1;            for(int j=0;j<N;j++)            {                if(i<kind[j][0]) continue;                int s=dp[i-kind[j][0]]+kind[j][1];                if(!vis[i-kind[j][0]][j]) s+=kind[j][2];                if(dp[i]<s)                {                    dp[i]=s;                    flag=j;                }            }            if(flag>=0)            {                for(int j=0;j<N;j++)                {                    vis[i][j]=vis[i-kind[flag][0]][j];                }                vis[i][flag]++;            }        }        int tmp=0;        for(int i=1;i<=M;i++)           tmp=max(tmp,dp[i]);        printf("%d\n",tmp);    }    return 0;}