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]>=0) return f[n][m];

    if (n==0) return 0;

    if (m<v[n])//fang bu liao
    {
        return dp(n-1,m);
    }
    else
    {
        f[n][m]=f[n-1][m];
        f[n][m]=max(dp(n-1,m),dp(n-1,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=1;i<=n;i++)
            scanf("%d",&w[i]);
        for (i=1;i<=n;i++)
            scanf("%d",&v[i]);

        memset(f,-1,sizeof(f));

            printf("%d\n",dp(n,m));
    }
    return 0;
}

 

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=1;i<=n;i++)
            scanf("%d",&w[i]);
        for (i=1;i<=n;i++)
            scanf("%d",&v[i]);

        for (j=0;j<=m;j++) f[0][j]=0;

        for (i=1;i<=n;i++)
            for (j=0;j<=m;j++)
            {
                f[i][j]=f[i-1][j];
                if (j>=v[i])
                    f[i][j]=max(f[i-1][j],f[i-1][j-v[i]]+w[i]);
            }
            printf("%d\n",f[n][m]);
    }
    return 0;
}

 

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=1;i<=n;i++)
            scanf("%d",&w[i]);
        for (i=1;i<=n;i++)
            scanf("%d",&v[i]);

        memset(f,0,sizeof(f));
        for (i=1;i<=n;i++)
            for (j=m;j>=0;j--)
            {
                if (j>=v[i])
                    f[j]=max(f[j],f[j-v[i]]+w[i]);
            }
            printf("%d\n",f[m]);
    }
    return 0;
}