• 洛谷——P2925 [USACO08DEC]干草出售Hay For Sale


    https://www.luogu.org/problem/show?pid=2925

    题目描述

    Farmer John suffered a terrible loss when giant Australian cockroaches ate the entirety of his hay inventory, leaving him with nothing to feed the cows. He hitched up his wagon with capacity C (1 <= C <= 50,000) cubic units and sauntered over to Farmer Don's to get some hay before the cows miss a meal.

    Farmer Don had a wide variety of H (1 <= H <= 5,000) hay bales for sale, each with its own volume (1 <= V_i <= C). Bales of hay, you know, are somewhat flexible and can be jammed into the oddest of spaces in a wagon.

    FJ carefully evaluates the volumes so that he can figure out the largest amount of hay he can purchase for his cows.

    Given the volume constraint and a list of bales to buy, what is the greatest volume of hay FJ can purchase? He can't purchase partial bales, of course. Each input line (after the first) lists a single bale FJ can buy.

    约翰遭受了重大的损失:蟑螂吃掉了他所有的干草,留下一群饥饿的牛.他乘着容量为C(1≤C≤50000)个单位的马车,去顿因家买一些干草. 顿因有H(1≤H≤5000)包干草,每一包都有它的体积Vi(l≤Vi≤C).约翰只能整包购买,

    他最多可以运回多少体积的干草呢?

    输入输出格式

    输入格式:

    • Line 1: Two space-separated integers: C and H

    • Lines 2..H+1: Each line describes the volume of a single bale: V_i

    输出格式:

    • Line 1: A single integer which is the greatest volume of hay FJ can purchase given the list of bales for sale and constraints.

    输入输出样例

    输入样例#1:
    7 3 
    2 
    6 
    5 
    
    输出样例#1:
    7 
    

    说明

    The wagon holds 7 volumetric units; three bales are offered for sale with volumes of 2, 6, and 5 units, respectively.

    Buying the two smaller bales fills the wagon.

     1 #include <algorithm>
     2 #include <cstdio>
     3 
     4 using namespace std;
     5 
     6 int n,m,w[50000];
     7 int f[500000];
     8 
     9 int main()
    10 {
    11     scanf("%d%d",&m,&n);
    12     for(int i=1;i<=n;i++)
    13         scanf("%d",&w[i]);
    14     for(int i=1;i<=n;i++)
    15         for(int j=m;j>=w[i];j--)
    16             if(f[j-w[i]]+w[i]>f[j])
    17                 f[j]=f[j-w[i]]+w[i];
    18     printf("%d",f[m]);
    19     return 0;
    20 }
    ——每当你想要放弃的时候,就想想是为了什么才一路坚持到现在。
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  • 原文地址:https://www.cnblogs.com/Shy-key/p/6718327.html
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