【POJ1157】LITTLE SHOP OF FLOWERS (简单动态规划)
2011-04-25 09:20
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该题也是经典的动态规划,题目叙述的依然很麻烦,其实简化一下就是这样的:例如下面这个例子就是:3表示行,5表示列,然后在下面的3行5列每一行选一个数,使这3个数最大,要求选的数列数必须依次增大,就是从左上方想右下方选3个数使和最大。
3 5
7 23 -5 -24 16
5 21 -4 10 23
-21
5 -4 -20 20
状态转移方程 arr[i][j]=max{arr[i-1][ 0...(j-1)]}+arr[i][j];
其中arr[][]存储的是当前最优值
3 5
7 23 -5 -24 16
5 21 -4 10 23
-21
5 -4 -20 20
状态转移方程 arr[i][j]=max{arr[i-1][ 0...(j-1)]}+arr[i][j];
其中arr[][]存储的是当前最优值
#include <iostream> using namespace std; #define maxlen 102 int arr[maxlen][maxlen]; int main(){ freopen("1.txt","r",stdin); int n,m,max; while(cin>>n>>m){ for (int i=0;i<n;i++) for (int j=0;j<m;j++) cin>>arr[i][j]; for (int i=1;i<n;i++){ for (int j=i;j<m;j++){ max=-99999; for (int k=0;k<j;k++) if(max<arr[i-1][k]) max=arr[i-1][k]; arr[i][j]+=max; } } max=-99999; for (int i=0;i<m;i++) if(max<arr[n-1][i]) max=arr[n-1][i]; cout<<max<<endl; } }
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