習DP,甚是迷惑,審其題,雖解其意,但無從下口,故棄而學揹包,今日,重拾之,遂發現其乃美味也。故知揹包乃DP之基也~
嘻嘻~~小小嘚瑟一下
我是剛開始接觸算法,就被諸位大神弄去學了DP,但是一知半解,遇到這道題的時候,怎麼也想不出該怎麼做。在我的師兄們(比我高一級的小菜,噓,別讓他聽到了)的指引下,看了揹包,今天再來看這道題~~~~大神 你在逗我玩嗎~~~~~
建議剛開始學dp的人 , 一定要去看看揹包九講,很精彩~~~
廢話少數,切入正題
Description
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?
Input
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.
Output
Sample Input
Sample Output
題意分析:一個骨頭蒐集愛好者,收集各類各樣的骨頭,這個骨頭愛好者有一個大袋子,容積是一定的,現給出一定的骨頭,每個骨頭的容積和價值是固定的,現在我們往袋子中裝入骨頭,求袋子容積所能承受的條件下的可裝的最大價值。
思路分析:這是個典型的01揹包,狀態轉移方程模板如下
for(int i = 1; i <= N; i++)//N骨頭的總數
{
for(int j = V; j <= v[i]; j--)//本題中,V指的是袋子的容積,v[i]指第i個骨頭的體積,d[i]指第i個骨頭的價值
{
dp[j]=max(dp[j],dp[j-v[i]]+d[i]);
}
}
現給出AC代碼
#include <iostream>
#include<cstdio>
#include<string.h>
using namespace std;
const int MAX = 1000+10;
int dp[MAX];
int d[MAX];
int v[MAX];
int T,N,V;
int main()
{
scanf("%d",&T);
while(T--)
{
memset(dp,0,sizeof(dp));
scanf("%d%d",&N,&V);
for(int i = 1; i <= N; i++)
{
scanf("%d",&d[i]);
}
for(int j = 1; j <= N; j++)
{
scanf("%d",&v[j]);
}
for(int i = 1; i <= N; i++)
{
for(int j = V; j >= v[i];j--)
{
dp[j]=max(dp[j],dp[j-v[i]]+d[i]);
}
}
printf("%d\n",dp[V]);
}
return 0;
}