poj2602 bone 0-1背包

 

          Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14336    Accepted Submission(s): 5688

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

ACcode:

#include <iostream>#include<cstdio>#include<string.h>using namespace std;int value[1001],volume[1001],dp[1001];int  Max(int  x,int  y){    return (x>y)?x:y;}int main(){   int t,n,V;  scanf("%d",&t);   while(t--)   {       scanf("%d%d",&n,&V);       for(int i=0;i<n;i++)        scanf("%d",&value[i]);        for(int i=0;i<n;i++)            scanf("%d",&volume[i]);            memset(dp,0,sizeof(dp));        for(int i=0;i<n;i++)            for(int v=V;v>=volume[i];v--)        dp[v]=Max(dp[v],dp[v-volume[i]]+value[i]);//0-1背包，要么不装，要么装        printf("%d",dp[V]);   }    return 0;}<span style="BACKGROUND-COLOR: #ffffff">总结：</span>

<span style="BACKGROUND-COLOR: #ffffff">0-1背包伪代码：</span>

<p>for i=1..N</p><p>    for v=V..0</p><p>        dp[v]=max{dp[v],dp[v-c[i]]+w[i]};</p>