【杭电2602】01背包

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 51515    Accepted Submission(s): 21692

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
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){    int n,v[1002],t,vv;    int value[1102],rad[1102];    scanf("%d",&t);    while(t--)    {        memset(rad,0,sizeof(rad));        scanf("%d%d",&n,&vv);        for(int i=1;i<=n;i++)        scanf("%d",&value[i]);        for(int i=1;i<=n;i++)        scanf("%d",&v[i]);        for(int i=1;i<=n;i++)        {            for(int j=vv;j>=v[i];j--)            rad[j]=max(rad[j-v[i]]+value[i],rad[j]);        }        printf("%d\n",rad[vv]);    }    return 0;}

二维数组
//用子问题定义状态：即f[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。
//则其状态转移方程便是：
//f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}
123456789101112131415161718192021222324252627282930313233
#include <stdio.h>  #include <string.h>#include<algorithm>using namespace std;  int dp[1002][1002];  int main()  {      int i = 0, j = 0, t = 0, n = 0, v = 0;      int W[1002], V[1002];      scanf("%d",&t);      while(t--)      {          scanf("%d %d",&n, &v);          for(i = 0; i<n; i++)              scanf("%d",&W[i]); //价值          for(i = 0; i<n; i++)              scanf("%d",&V[i]);  //体积         memset(dp, 0, sizeof(dp));          for(i = 0; i<V[0]; i++)  //初值             dp[0][i] = 0;          for(i = V[0]; i<=v; i++)              dp[0][i] = W[0];          for(i = 1; i<n; i++)          {              for(j = 0; j<V[i]; j++)                  dp[i][j] = dp[i-1][j]; //比当前物品体积小，不加入              for(j = V[i]; j<=v; j++)                  dp[i][j] = max(dp[i-1][j],dp[i-1][j-V[i]]+W[i]);  //比当前体积大，选择加或不加         }          printf("%d\n",dp[n-1][v]);      }      return 0;  }