HDU - 2602 Bone Collector(01背包)

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 2 ³¹).

 

Sample Input

 15 101 2 3 4 55 4 3 2 1 

 

Sample Output

 14

题解：直接就是01背包，把前i个物品装到容器量为j的背包中的最大总重量

f(i,j) = max(f(i-1,j),f(i-1,j-v[i])+w[i])

N为数量，V为最大的体积

最终答案为f(N,V);

#include <iostream>#include <cstdio>using namespace std;int N,V;int v[1005];int w[1005];int dp[1005][1005];int main(){    int t;    scanf("%d",&t);    while(t--){        scanf("%d%d",&N,&V);        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 = 0; j <= V; j++){                dp[i][j] = (i==1? 0 : dp[i-1][j]);                if(j >= v[i]) dp[i][j] = max(dp[i][j],dp[i-1][j-v[i]]+w[i]);            }        printf("%d\n",dp[N][V]);    }}