ACM 之 M - 基础DP
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 2 ^31).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
理解:
第一道DP题，也就是01背包问题，我的做法就是多看多写多理解。
另外，个人感觉用一位数组更简洁明了。
代码部分
#include <cstdio>
#include <cstring>
using namespace std;
int va[101],vo[101],dp[100002],t,n,v;
inline int Max(int x,int y)
{return x>y?x:y;}
int main()
{
while(scanf("%d",&t)!=EOF)
{
getchar();
for(int i=0;i<t;i++)
{
scanf("%d%d",&va[i],&vo[i]);
}getchar();
scanf("%d",&v);getchar();
memset(dp,0,sizeof(dp));
for(int i=0;i<t;i++)
for(int j=vo[i];j<=v;j++)
dp[j]=Max(dp[j],dp[j-vo[i]]+va[i]);
printf("%d\n",dp[v]);
}
return 0;
}
