hdu 2602 背包问题之01背包
2014-02-15 12:37
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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 ?
![](http://acm.hdu.edu.cn/data/images/C154-1003-1.jpg)
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 231).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
这是我的第一个背包问题。纪念一下,啦啦啦~~
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)
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,
下面是我写的:
若用一位数组来解决的话
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 ?
![](http://acm.hdu.edu.cn/data/images/C154-1003-1.jpg)
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 231).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
这是我的第一个背包问题。纪念一下,啦啦啦~~
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 |
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 [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; }
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