[USACO08DEC]乾草出售Hay For Sale——動態規劃
題目來源
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：蟑螂喫掉了他所有的乾草，留下一羣飢餓的牛．他乘着容量爲C(1≤C≤50000)個單位的馬車，去頓因家買一些乾草． 頓因有H(1≤H≤5000)包乾草，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乾草呢？
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.
解題思路
本題很明顯可以用01揹包解決，但是如果直接套用01揹包的格式，會在最後一個測試點超時，因此我添加了部分優化：
- 用滾動數組f[i]表示當車輛容量爲i時能裝乾草的最大體積
- 由於需要從小到大遍歷車的容量，假設當前車的容量是a，那麼f[a]一定小於等於a。（證明 : 車輛最多能裝體積與車輛容量相等的乾草）這說明當f[a]取a時就不用再一次對f[a]進行更新了，因爲此時的f[a]已經取到最大值了
- 由於車的總容量爲c，那麼這輛車最多隻能裝體積爲c的乾草。因此如果已經可以裝到體積爲c的乾草就直接輸出結果並退出
源代碼
#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體積的乾草
cout << c;//能夠裝滿
return 0;//退出
}
}
cout << f[c];
return 0;
}