CRB and His Birthday(背包)
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CRB and His Birthday
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 357 Accepted Submission(s): 191
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
题意:今天是CRB的生日,他的妈妈去商店给他买礼物,由于收银员是他妈妈的好朋友,所以收银员会按照不同礼 物的件数x赠与CRB的妈妈(a*x+b)块糖果。CRB的妈妈总共带了w元钱,总共有n种礼物。t组输入,每组先输入w 和n,接下来n行每行是该种礼物需要花费的价格和对应的a与b。求CRB的妈妈最多能获得多少糖果。
思路:因为每件物品都是无穷的所以是完全背包,但是只有在买第一件物品是多加b,所以将物品分成两部分,一部分是只有一件物品,进行01背包,一部分进行完全背包
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <queue>#include <map>#include <algorithm>#define INF 0x3f3f3f3fusing namespace std;typedef unsigned long long LL;int Dp[2100];int n,m;int main(){ int T; int w,a,b; scanf("%d",&T); while(T--) { scanf("%d %d",&m,&n); memset(Dp,0,sizeof(Dp)); for(int i=1;i<=n;i++) { scanf("%d %d %d",&w,&a,&b); for(int j=m;j>=w;j--)//01背包 { Dp[j]=max(Dp[j],Dp[j-w]+a+b); } for(int j=w;j<=m;j++)//完全背包 { Dp[j]=max(Dp[j],Dp[j-w]+a); } } printf("%d\n",Dp[m]); } return 0;}
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