日韩性视频-久久久蜜桃-www中文字幕-在线中文字幕av-亚洲欧美一区二区三区四区-撸久久-香蕉视频一区-久久无码精品丰满人妻-国产高潮av-激情福利社-日韩av网址大全-国产精品久久999-日本五十路在线-性欧美在线-久久99精品波多结衣一区-男女午夜免费视频-黑人极品ⅴideos精品欧美棵-人人妻人人澡人人爽精品欧美一区-日韩一区在线看-欧美a级在线免费观看

歡迎訪問 生活随笔!

生活随笔

當前位置: 首頁 > 编程资源 > 编程问答 >内容正文

编程问答

HDU2602-Bone Collector

發布時間:2025/7/25 编程问答 26 豆豆
生活随笔 收集整理的這篇文章主要介紹了 HDU2602-Bone Collector 小編覺得挺不錯的,現在分享給大家,幫大家做個參考.

描述:

  Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …?

  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 ??

  The first line contain a integer T , the number of cases.?

  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.

  One integer per line representing the maximum of the total value (this number will be less than 2?31).

代碼:

  最基本的01背包。

#include<stdio.h> #include<string.h> #include<iostream> #include<stdlib.h> #include <math.h> using namespace std; #define N 1005int main(){int T,n,v;int worth[N],cost[N],dp[N];scanf("%d",&T);while( T-- ){scanf("%d%d",&n,&v);for( int i=1;i<=n;i++ )scanf("%d",&worth[i]);for( int i=1;i<=n;i++ )scanf("%d",&cost[i]);memset(dp,0,sizeof(dp));for( int i=1;i<=n;i++ ){for( int j=v;j>=cost[i];j-- ){dp[j]=max(dp[j],dp[j-cost[i]]+worth[i]);}}printf("%d\n",dp[v]);}system("pause");return 0; }

?

轉載于:https://www.cnblogs.com/lucio_yz/p/4748942.html

總結

以上是生活随笔為你收集整理的HDU2602-Bone Collector的全部內容,希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯,歡迎將生活随笔推薦給好友。