洛谷P2925 [USACO08DEC]干草出售Hay For Sale

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.

01背包。常数看着挺大但是不会T

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 #include<vector>
 8 using namespace std;
 9 int read(){
10     int x=0,f=1;char ch=getchar();
11     while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
12     while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
13     return x*f;
14 }
15 int f[50010];
16 int n,c,v[5010];
17 int main(){
18     c=read();n=read();
19     int i,j;
20     for(i=1;i<=n;i++)v[i]=read();
21     f[0]=1;
22     for(i=1;i<=n;i++){
23         for(j=c;j>=v[i];j--){
24             f[j]|=f[j-v[i]];
25         }
26     }
27     for(j=c;j>=0;j--){
28         if(f[j]){
29             printf("%d
",j);
30             break;
31         }
32     }
33     return 0;
34 }
原文地址:https://www.cnblogs.com/SilverNebula/p/6050402.html