M - 基础DP

M - 基础DP

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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 2 ³¹).

Sample Input

1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output

14

//第一行代表有T个测试案例
第二行 n，m 代表有 n 个物品，背包容量为 m 。接下来两行分别是 n 个物品的价值，体积

//动态规划入门，看了很久才懂。
我先用的递归做的 5696kb 452ms

 #include <stdio.h>
 #include <string.h>

 #define MAX 1005

 int v[MAX];
 int w[MAX];
 int f[MAX][MAX];

 int max(int a,int b)
 {
     return a>b?a:b;
 }

 int dp(int n,int m)
 {
     if (f[n][m]>=) return f[n][m];

     if (n==) return ;

     if (m<v[n])//fang bu liao
     {
         return dp(n-,m);
     }
     else
     {
         f[n][m]=f[n-][m];
         f[n][m]=max(dp(n-,m),dp(n-,m-v[n])+w[n]);
     }
     return f[n][m];
 }

 int main()
 {
     int i,t,n,m;
     scanf("%d",&t);
     while (t--)
     {
         scanf("%d%d",&n,&m);
         for (i=;i<=n;i++)
             scanf("%d",&w[i]);
         for (i=;i<=n;i++)
             scanf("%d",&v[i]);

         memset(f,-,sizeof(f));

             printf("%d\n",dp(n,m));
     }
     return ;
 }

递推 5592kb 78ms

 #include <stdio.h>

 #define MAX 1005

 int v[MAX];
 int w[MAX];
 int f[MAX][MAX];

 int max(int a,int b)
 {
     return a>b?a:b;
 }

 int main()
 {
     int i,j,t,n,m;
     scanf("%d",&t);
     while (t--)
     {
         scanf("%d%d",&n,&m);
         for (i=;i<=n;i++)
             scanf("%d",&w[i]);
         for (i=;i<=n;i++)
             scanf("%d",&v[i]);

         for (j=;j<=m;j++) f[][j]=;

         for (i=;i<=n;i++)
             for (j=;j<=m;j++)
             {
                 f[i][j]=f[i-][j];
                 if (j>=v[i])
                     f[i][j]=max(f[i-][j],f[i-][j-v[i]]+w[i]);
             }
             printf("%d\n",f[n][m]);
     }
     return ;
 }

一维数组 1784kb 15ms

 #include <stdio.h>
 #include <string.h>

 #define MAX 1005

 int v[MAX];
 int w[MAX];
 int f[MAX];

 int max(int a,int b)
 {
     return a>b?a:b;
 }

 int main()
 {
     int i,j,t,n,m;
     scanf("%d",&t);
     while (t--)
     {
         scanf("%d%d",&n,&m);
         for (i=;i<=n;i++)
             scanf("%d",&w[i]);
         for (i=;i<=n;i++)
             scanf("%d",&v[i]);

         memset(f,,sizeof(f));
         for (i=;i<=n;i++)
             for (j=m;j>=;j--)
             {
                 if (j>=v[i])
                     f[j]=max(f[j],f[j-v[i]]+w[i]);
             }
             printf("%d\n",f[m]);
     }
     return ;
 }