洛谷 ——2925干草出售Hay For Sale

最新推荐文章于 2022-09-08 00:08:38 发布

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最新推荐文章于 2022-09-08 00:08:38 发布

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原文链接：http://www.cnblogs.com/z360/p/6719157.html

面对巨型澳大利亚蟑螂吃光所有干草的困境，农夫约翰利用容量为C的马车前往农夫唐处购买干草。本篇介绍了如何通过算法帮助约翰确定在不同体积的干草捆中能购买的最大总体积。

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：

输出样例#1：

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.

代码

#include <algorithm>
#include <cstdio>

using namespace std;

int n,v,w[50000];
int f[500000];

int main()
{
    scanf("%d%d",&v,&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&w[i]);
    for(int i=1;i<=n;i++)
        for(int m=v;m>=w[i];m--)
            if(f[m-w[i]]+w[i]>f[m])
                f[m]=f[m-w[i]]+w[i];
    printf("%d",f[v]);
    return 0;
}