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 ?
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.
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 ;
}