[USACO08DEC]干草出售Hay For Sale

题目

Description

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？

Input

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

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

Output

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

Sample Input

7 3 
2 
6 
5 

Sample Output

7 

---

思路

很明显是一道背包题；

但是直接背包模板是过不了的；会$TLE$;

只需要一个特判就ok了，如果已经背出了背包的最大容量，那么直接输出最大容量，然后结束程序；

由于博主蒟蒻，非常喜欢快，所以就加了$bitset$ 优化；

 

代码

#include<bits/stdc++.h>
#define re register
typedef long long ll;
using namespace std;
inline ll read()
{
    ll a=0,f=1; char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();}
    while (c>='0'&&c<='9') {a=a*10+c-'0'; c=getchar();}
    return a*f;
}
ll n,m;
ll a[6010];
bitset<50010> dp;//bitset的定义
int main()
{
    dp[0]=1;//初始化，没它，你wa
    m=read(); n=read();
    for(re ll i=1;i<=n;i++)
    {
        a[i]=read();
        dp|=(dp<<a[i]);
        if(dp[m])//特判
        {
            printf("%lld\n",m);
            return 0;
        }
    }
    for(re ll i=m;i>=1;i--)
    if(dp[i])//逆循环，找到最大的权值
    {
        printf("%lld\n",i);
        break;
    }
        //return 0；
}