HDU 2602 Bone Collector （01背包）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 65956    Accepted Submission(s): 27502

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

经典01背包问题，二维状态转移方程：dp[i][j] = max(dp[i-1][j],dp[i-1][j-c[i]]+w[i]),一维状态转移方程：

dp[i] = max(dp[i],dp[i-c[i]]+w[i]);  对于每一种物品都是放或者不放。

裸题，代码（用的一维数组，比较省时）：

#include<cstdio>#include<iostream>#include<algorithm>#include<queue>#include<stack>#include<cstring>#include<string>#include<vector>#include<map>using namespace std;#define mem(a,b) memset(a,b,sizeof(a))typedef long long ll;const int maxn = 1e3+5;const int ff = 0x3f3f3f3f;int n,m;int dp[maxn],w[maxn],v[maxn];int main(){int t;cin>>t;while(t--){mem(v,0); mem(w,0);mem(dp,0);scanf("%d %d",&n,&m);for(int i = 1;i<= n;i++)scanf("%d",&w[i]);for(int i = 1;i<= n;i++)scanf("%d",&v[i]);for(int i = 1;i<= n;i++)for(int j = m;j>= v[i];j--)dp[j] = max(dp[j],dp[j-v[i]]+w[i]);printf("%d\n",dp[m]);}return 0;}