hdu 2602 背包问题之01背包
http://acm.hdu.edu.cn/showproblem.php?pid=2602
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 ?
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
5 10
1 2 3 4 5
5 4 3 2 1
横向表示w（空间，从0开始），竖向表示物品数（从1开始）。对于这个示例数据；我的程序模拟演示如下，请看表格：（列之所以从零开始，是因为骨头的重量可能为零，至于为什么，就留给读者了）
解释：比如，第一行，第一列表示：现在考虑第一个物品（1，5）对于空间为零（列号表示空间大小）不能装下，所以f[1][0]=0,而第一行第二列空间就变成了1，可以装下第一个物品，故f[1][1]=5。再如，第二行第四列，对于第二个物品（2,4）在装了第一个后还剩空间为2，正好装下，故f[2][3]=5+4;以此类推……（看不懂自己画画表格再模拟一下）
| 0 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 |
| 0 | 5 | 5 | 9 | 9 | 9 | 9 | 9 | 9 | 9 | 9 |
| 0 | 5 | 5 | 9 | 9 | 9 | 12 | 12 | 12 | 12 | 12 |
| 0 | 5 | 5 | 9 | 9 | 9 | 12 | 12 | 12 | 12 | 14 |
| 0 | 5 | 5 | 9 | 9 | 9 | 12 | 12 | 12 | 12 | 14 |
如果你在自己写代码，试试这组数据，这个过了，基本上就AC了，
1 5 0 2 4 1 5 1 0 0 1 0 0 答案是12下面是我写的：
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
int f[1005][1005],w[1005],v[1005];
int main()
{
int m;
scanf("%d",&m);
while(m--)
{
memset(f,0,sizeof(f));// 先清零是很必要的
memset(w,0,sizeof(w));
memset(v,0,sizeof(v));
int n,V;
scanf("%d%d",&n,&V);
for(int i=1;i<=n;i++)
cin >> v[i];
for(int i=1;i<=n;i++)
cin >> w[i];
for(int i=1;i<=n;i++)
for(int j=0;j<=V;j++)
{
f[i][j]=f[i-1][j];
if(j>=w[i]&&f[i][j]<f[i-1][j-w[i]]+v[i])
f[i][j]=f[i-1][j-w[i]]+v[i];
}
cout <<f[n][V]<< endl;
}
return 0;
}
若用一位数组来解决的话
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
int f[1005],w[1005],v[1005];
int main()
{
int m;
scanf("%d",&m);
while(m--)
{
memset(f,0,sizeof(f));// 先清零是很必要的
memset(w,0,sizeof(w));
memset(v,0,sizeof(v));
int n,V;
scanf("%d%d",&n,&V);
for(int i=1;i<=n;i++)
cin >> v[i];
for(int i=1;i<=n;i++)
cin >> w[i];
for(int i=1;i<=n;i++)
{
for(int j=V;j>=w[i];j--)
{
f[j]=max(f[j],f[j-w[i]]+v[i]);
//cout << f[j] << " ";
}
// cout << endl;
}
// for(int i=0;i<=V;i++)
// cout <<f[i]<< endl;
cout << f[V]<< endl;
}
return 0;
}