HDU 2602 (0-1背包)

                                     Bone Collector

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

分析：简单01背包问题，dp(i,j)表示前i件物品放进背包大小为j的最大值，状态方程dp(i,j)=max(dp(i-1,j),dp(i-1,j-a[i])+value[i]);

AC代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=1000+10;int value[maxn];int dp[maxn][maxn];int a[maxn];int main(){int T;scanf("%d",&T);while(T--){int N,V;scanf("%d%d",&N,&V);for(int i=1;i<=N;i++)  scanf("%d",&value[i]);for(int i=1;i<=N;i++)  scanf("%d",&a[i]); memset(dp,0,sizeof(dp)); for(int i=1;i<=N;i++){ for(int j=0;j<=V;j++){ dp[i][j]=dp[i-1][j]; if(a[i]<=j)dp[i][j]=max(dp[i][j],dp[i-1][j-a[i]]+value[i]); } } printf("%d\n",dp[N][V]);}return 0;} 

对空间的优化（二维化一维）

AC代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=1000+10;int value[maxn];int dp[maxn];int a[maxn];int main(){int T;scanf("%d",&T);while(T--){int N,V;scanf("%d%d",&N,&V);for(int i=1;i<=N;i++)  scanf("%d",&value[i]);for(int i=1;i<=N;i++)  scanf("%d",&a[i]); memset(dp,0,sizeof(dp)); for(int i=1;i<=N;i++){ for(int j=V;j>=0;j--){ if(j>=a[i])dp[j]=max(dp[j],dp[j-a[i]]+value[i]); } } printf("%d\n",dp[V]);}return 0;}