洛谷P2925 [USACO08DEC]干草出售Hay For Sale
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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
- 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
7 3 2 6 5
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.
#include<iostream>#include<algorithm>#include<cstdio>#define MAXN 10010using 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;}
- P2925 [USACO08DEC]干草出售Hay For Sale
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- 洛谷 P2925 [USACO08DEC]干草出售Hay For Sale
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