洛谷P2925 [USACO08DEC]干草出售Hay For Sale-优快云博客

面对巨型澳大利亚蟑螂吞噬全部干草的灾难，农夫John需要利用有限容量的马车从农夫Don处购买尽可能多的干草。本篇介绍了一个01背包问题的应用案例，详细解释了如何通过算法确定能装载的最大干草体积。

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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.

01背包，套模板。

但朴素01背包会 TLE ，那么加一个特判：当体积达到 C 时，跳出。

附代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#define MAXN 10010
using namespace std;
int n,m,w[MAXN],f[MAXN*10];
inline int read(){
	int date=0,w=1;char c=0;
	while(c<'0'||c>'9'){if(c=='-')w=-1;c=getchar();}
	while(c>='0'&&c<='9'){date=date*10+c-'0';c=getchar();}
	return date*w;
}
int main(){
	m=read();n=read();
	for(int i=1;i<=n;i++)w[i]=read();
	for(int i=1;i<=n;i++)
		for(int j=m;j>=w[i];j--){
			f[j]=max(f[j],f[j-w[i]]+w[i]);
			if(f[j]==m){
				printf("%d\n",m);
				return 0;
			}
		}
	printf("%d\n",f[m]);
	return 0;
}