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

2024-08-11 19:57:42   221人阅读

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
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
 
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；
 
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[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;}

 

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


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