P2925 [USACO08DEC]幹草出售Hay For Sale
題目描述
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.
約翰遭受了重大的損失：蟑螂吃掉了他所有的幹草，留下一群饑餓的牛．他乘著容量為C(1≤C≤50000)個單位的馬車，去頓因家買一些幹草． 頓因有H(1≤H≤5000)包幹草，每一包都有它的體積Vi(l≤Vi≤C).約翰只能整包購買，
他最多可以運回多少體積的幹草呢？
輸入輸出格式
輸入格式：
-
Line 1: Two space-separated integers: C and H
- Lines 2..H+1: Each line describes the volume of a single bale: V_i
輸出格式：
- Line 1: A single integer which is the greatest volume of hay FJ can purchase given the list of bales for sale and constraints.
輸入輸出樣例
輸入樣例#1：7 3 2 6 5輸出樣例#1：
7
說明
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.
竟然超時：
#include<iostream>
#include<cstdio>
using namespace std;
int f[50009],v[5009],n,m;
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++)
scanf("%d",&v[i]);
for(int i=1;i<=m;i++)
for(int j=n;j>=v[i];j--)
if(f[j]<f[j-v[i]]+v[i])
f[j]=f[j-v[i]]+v[i];
printf("%d",f[n]);
return 0;
}
#include<iostream>
using namespace std;
int c,h;//c容量 h種情況
int f[50005];
int v[50005];
int main()
{
cin >> c >> h;
for(int i = 1;i <= h;i++)
cin >> v[i];
for(int i = 1;i <= h;i++)
{
for(int a = c;a >= v[i];a--)
{
if(f[a] == a)
continue; //此時f[a]已經取到最大值 就不用再對f[a]進行更新
if(f[a - v[i]] + v[i] > f[a])
f[a] = f[a - v[i]] + v[i];
}
if(f[c] == c)
{//判斷是否已經能夠裝滿c體積的幹草
cout << c;//能夠裝滿
return 0;//退出
}
}
cout << f[c];
return 0;
}
P2925 [USACO08DEC]幹草出售Hay For Sale