HDU 5410 CRB and His Birthday

题目链接：http://acm.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

1 100 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.

题目的大概意思是有M元钱，有N件物品，没件物品的花费是W[i]，第i件物品买X件能得到A[i]*X+B[i]块糖，没件物品买的数量没有上限

问这M元钱最多能得到多少块糖。

问题分析：没种物品买第一件能得到A[i]+B[i]块糖，以后每买一件只能多得到A[i]块糖，也就是每种物品买第一件比买后面的得到的糖多，

根据贪心的思想很容易想到要先每种物品买一件再买大于一件，这样就把问题分成了两个阶段，第一阶段每种物品最多只买一件，

0/1背包，第二阶段，没种物品能买无限件，完全背包。

 1 #include <iostream>
 2 #include <cstring>
 3 using namespace std;
 4
 5 int ma(int a,int b)
 6 {
 7     if (a>b)
 8     return a;
 9     else
10     return b;
11 }
12 int main()
13 {
14     int ans[12000],v,a[1100],b[1100],c[1100],n,m,T;
15     cin>>T;
16     while (T--)
17     {
18           cin>>v>>m;
19           for (int i=1;i<=m;i++)
20           cin>>c[i]>>a[i]>>b[i];
21           memset(ans,0,sizeof(ans));
22           for (int i=1;i<=m;i++)//处理第一阶段，0/1背包
23           {
24               for (int j=v;j>=c[i];j--)
25               {
26                   ans[j]=ma(ans[j],ans[j-c[i]]+a[i]+b[i]);
27               }
28           }
29           for (int i=1;i<=m;i++)//处理第二阶段，完全背包。
30           {
31               for (int j=c[i];j<=v;j++)
32               {
33                    ans[j]=ma(ans[j],ans[j-c[i]]+a[i]);
34               }
35           }
36           cout <<ans[v]<<endl;
37     }
38     return 0;
39 }

View Code

转载于:https://www.cnblogs.com/arno-my-boke/p/4763998.html

HDU 5410 CRB and His Birthday相关推荐

HDU 5410 CRB and His Birthday ——（完全背包变形）
对于每个物品,如果购买,价值为A[i]*x+B[i]的背包问题. 先写了一发是WA的= =.代码如下: 1 #include <stdio.h> 2 #include <algori ...
hdu 5410(背包问题变形)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5410 解题思路:令dp[i][j][0]表示前i种物品,共j钱,不买第i种物品所能买到的最大值.dp[ ...
HDU - 5411 CRB and Puzzle 矩阵快速幂
HDU - 5411 考虑直接dp会T, 用矩阵优化一下就好了. #include<bits/stdc++.h> #define LL long long #define LD long ...
hdu 5411 CRB and Puzzle（矩阵快速幂）
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5411 解题思路: 题目大意: 给定n个点常数m 下面n行第i行第一个数字表示i点的出边数,后面给出 ...
HDU 5411 CRB and Puzzle （2015年多校比赛第10场）
1.题目描写叙述:点击打开链接 2.解题思路:本题实际是是已知一张无向图.问长度小于等于m的路径一共同拥有多少条. 能够通过建立转移矩阵利用矩阵高速幂解决.当中,转移矩阵就是输入时候的邻接矩阵,同一时 ...
hdu 5411 CRB and Puzzle 矩阵高速幂
链接题解链接:http://www.cygmasot.com/index.php/2015/08/20/hdu_5411/ 给定n个点常数m 以下n行第i行第一个数字表示i点的出边数.后面给出这些 ...
hdu 5411 CRB and Puzzle
Sample Input 1 3 2 1 2 1 3 0 Sample Output 6 Hint possible patterns are ∅, 1, 2, 3, 1→2, 2→3 解释样例: 第 ...
【矩阵快速幂】 HDU 5411 CRB and Puzzle 等比
点击打开链接题意显然是求 A+A^2+A^3+....+A^m 这就是经典题目矩阵乘法十种经典题目递归解决后半部分提取 A^(m/2) 该题再特判下 m==1的时候 #include < ...
小算法小心情：背包问题就是陪你看花开向阳
一开始接触背包问题,总会有些困扰,无法完全理解,或者说不断地忘记所谓的公式,特神说分享,可以让许多人少走弯路,我觉得颇有道理,其实我想做的,只是努力让让算法变得看上去简单易懂一些,因为我也曾是个迷惘无 ...

HDU 5410 CRB and His Birthday

HDU 5410 CRB and His Birthday相关推荐

最新文章

热门文章