目錄
題目描述
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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 ?
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
題意分析
題意:一個喜歡收集骨頭的人,現在給你N個骨頭的價值以及其所佔的體積,
現在它有容量爲M的口袋,問能夠裝入的最大價值是多少。
典型的01揹包問題。數據中會給出一些 有價值但體積爲0的數據,謹須防範
AC代碼
一維數組寫法
#include <iostream>
#include <algorithm>
#include <memory.h>
#define MAXN 10005
using namespace std;
struct infor
{
int value;
int volume;
}x[MAXN];
int DP[MAXN];
int main()
{
int t,n,m,i,j;
cin>>t;
while(t--)
{
cin>>n>>m;
memset(DP,0,sizeof(DP));
for(i=0;i<n;i++)
{
cin>>x[i].value;
}
for(i=0;i<n;i++)
{
cin>>x[i].volume;
}
for(i=0;i<n;i++)
{
for(j=m;j>= x[i].volume; j--)
{
DP[j] = max( DP[j-x[i].volume ]+x[i].value , DP[j] );
}
}
cout<<DP[m]<<endl;
}
return 0;
}
二維數組寫法
#include <iostream>
#include <algorithm>
#include <memory.h>
#define MAXN 1005
using namespace std;
struct infor
{
int value;
int volume;
}x[MAXN];
int DP[MAXN][MAXN];
int main()
{
int t,n,m,i,j;
cin>>t;
while(t--)
{
cin>>n>>m;
memset(DP,0,sizeof(DP));
for(i=1;i<=n;i++)
{
cin>>x[i].value;
}
for(i=1;i<=n;i++)
{
cin>>x[i].volume;
}
for(i=1;i<=n;i++)
{
for(j=0; j<=m; j++ )
{
DP[i][j] = max( DP[i-1][ j-x[i].volume ] + x[i].value , DP[i-1][j] );
if(x[i].volume<=j && DP[i-1][j]<DP[i-1][j-x[i].volume]+x[i].value)
DP[i][j]=DP[i-1][j-x[i].volume]+x[i].value;
else
DP[i][j]=DP[i-1][j];
}
}
cout<<DP[n][m]<<endl;
}
return 0;
}