Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N(1 ≤ N≤ 3,402) available charms. Each charm iin the supplied list has a weight Wi(1 ≤ Wi≤ 400), a 'desirability' factor Di(1 ≤ Di≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M(1 ≤ M≤ 12,880).
Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.
Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: W i and D i输出Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints样例输入
4 6 1 4 2 6 3 12 2 7样例输出
23
经典的01背包问题,但是使用二维数组会WA
#include<cstdio>
#include<cstring>
#define max(x,y) (x)>(y)?(x):(y)
using namespace std;
int n,m;
int value[3500];
int weight[3500];
int matrix[3500][3500];
int main()
{
scanf("%d%d",&n,&m);
memset(matrix,0,sizeof(matrix));
value[0]=weight[0]=0;
for(int i=1;i<=n;i++)
{
scanf("%d%d",&weight[i],&value[i]);
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if(j>=weight[i])
matrix[i][j]=max(matrix[i-1][j],matrix[i-1][j-weight[i]]+value[i]);
else
matrix[i][j]=matrix[i-1][j];
}
}
printf("%d",matrix[n][m]);
return 0;
}
优化如下(来源:http://www.cnblogs.com/ECJTUACM-873284962/p/6815610.html)
可以进一步优化内存使用。上面计算f[i][j]可以看出,在计算f[i][j]时只使用了f[i-1][0……j],没有使用其他子问题,因此在存储子问题的解时,只存储f[i-1]子问题的解即可。这样可以用两个一维数组解决,一个存储子问题,一个存储正在解决的子问题。
再进一步思考,计算f[i][j]时只使用了f[i-1][0……j],没有使用f[i-1][j+1]这样的话,我们先计算j的循环时,让j=M……1,只使用一个一维数组即可。
for i=1……N
for j=M……1
f[j]=max(f[j],f[j-weight[i]+value[i])
AC代码如下:
#include<iostream>
using namespace std;
int f[35000];//全局变量,自动初始化为0
int weight[35000];
int value[35000];
#define max(x,y) (x)>(y)?(x):(y)
int main()
{
int N,M;
cin>>N;//物品个数
cin>>M;//背包容量
for (int i=1;i<=N; i++)
{
cin>>weight[i]>>value[i];
}
for (int i=1; i<=N; i++)
for (int j=M; j>=1; j--)
{
if (weight[i]<=j)
{
f[j]=max(f[j],f[j-weight[i]]+value[i]);
}
}
cout<<f[M]<<endl;//输出最优解
}