zoukankan      html  css  js  c++  java
  • Buns(dp+多重背包)

    C. Buns
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Lavrenty, a baker, is going to make several buns with stuffings and sell them.

    Lavrenty has n grams of dough as well as m different stuffing types. The stuffing types are numerated from 1 tom. Lavrenty knows that he has ai grams left of the i-th stuffing. It takes exactly bi grams of stuffing i and cigrams of dough to cook a bun with the i-th stuffing. Such bun can be sold for di tugriks.

    Also he can make buns without stuffings. Each of such buns requires c0 grams of dough and it can be sold ford0 tugriks. So Lavrenty can cook any number of buns with different stuffings or without it unless he runs out of dough and the stuffings. Lavrenty throws away all excess material left after baking.

    Find the maximum number of tugriks Lavrenty can earn.

    Input

    The first line contains 4 integers n, m, c0 and d0 (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10, 1 ≤ c0, d0 ≤ 100). Each of the following m lines contains 4 integers. The i-th line contains numbers ai, bi, ci and di (1 ≤ ai, bi, ci, di ≤ 100).

    Output

    Print the only number — the maximum number of tugriks Lavrenty can earn.

    Examples
    input
    10 2 2 1 7 3 2 100 12 3 1 10
    output
    241
    input
    100 1 25 50 15 5 20 10
    output
    200
    Note

    To get the maximum number of tugriks in the first sample, you need to cook 2 buns with stuffing 1, 4 buns with stuffing 2 and a bun without any stuffing.

    In the second sample Lavrenty should cook 4 buns without stuffings.

    思维:让你做蛋糕,有各种口味的奶油,和面,消耗一定的面一定口味的奶油可以卖d元,问最多买多少钱;因为是蛋糕肯定是整个整个的,所以要用到背包;多重背包;

    dp[i][k]代表前i种蛋糕在面小于等于k的最大利润;

    代码:

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    #include<map>
    #include<string>
    using namespace std;
    const int INF=0x3f3f3f3f;
    #define SI(x) scanf("%d",&x)
    #define PI(x) printf("%d",x)
    #define P_ printf(" ")
    #define mem(x,y) memset(x,y,sizeof(x))
    struct Node{
        int a,b,c,d;
    };
    Node dt[15];
    int dp[15][1010];
    int main(){
        int n;
        int m;
        while(~scanf("%d%d%d%d",&n,&m,&dt[0].c,&dt[0].d)){
            dt[0].a=100010;dt[0].b=1;
            for(int i=1;i<=m;i++)scanf("%d%d%d%d",&dt[i].a,&dt[i].b,&dt[i].c,&dt[i].d);
            mem(dp,0);
            for(int i=0;i<=m;i++){
                for(int j=0;j*dt[i].b<=dt[i].a;j++){
                    for(int k=n;k>=j*dt[i].c;k--){
                        if(i)dp[i][k]=max(dp[i][k],dp[i-1][k-j*dt[i].c]+j*dt[i].d);
                        else dp[i][k]=max(dp[i][k],dp[i][k-j*dt[i].c]+j*dt[i].d);
                    }    
                }
            }
            printf("%d
    ",dp[m][n]);
        }
        return 0;
    }
  • 相关阅读:
    linux命令: mount
    梳理一下uboot是如何从nandflash挂载文件系统的
    MDK的优化应用
    面向对象设计思想:面向对象设计的基本原则
    问题
    nodejs安装不了和npm安装不了的解决方法
    []: secureCRT连接ubuntu问题- The remote system refused the connection
    字符设备驱动[深入]:linux cdev详解
    使用MDK将STM32的标准库编译成lib使用
    liteos任务(二)
  • 原文地址:https://www.cnblogs.com/handsomecui/p/5255180.html
Copyright © 2011-2022 走看看