題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=2602


Problem Description 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 ?


Input 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.
Output One integer per line representing the maximum of the total value (this number will be less than 2³¹).
Sample Input 1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output 14

狀態(tài)轉(zhuǎn)移方程為：dp[i]=max(dp[i], dp[i-v[j]]+w[j]);

代碼如下：

#include<cstdio> #include<cstring> #include<iostream> using namespace std;int main(){int v[1001],w[1001],dp[1001],t;scanf("%d",&t);while(t--){int p,n;memset(dp,0,sizeof(dp));scanf("%d%d", &n, &p);for(int i=0; i<n; ++i){scanf("%d",&w[i]);}for(int i=0; i<n; ++i){scanf("%d",&v[i]);}for(int j=0; j<n; ++j){for(int i=p; i>=v[j]; --i){//dp[i]表示將前j件物品放進(jìn)容量為i的背包中的最大總價值 dp[i]=max(dp[i], dp[i-v[j]]+w[j]);}}printf("%d\n",dp[p]);}return 0; }

總結(jié)

以上是生活随笔為你收集整理的HD 2602 Bone Collector （0-1背包）的全部內(nèi)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(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。