openjudge-NOI 2.6-2728 摘花生

时间:2023-03-08 23:14:03
openjudge-NOI 2.6-2728 摘花生

题目链接:http://noi.openjudge.cn/ch0206/2728/

题解:
  某一个点只能从其左边或者上边走过来

  f[i][j]存储(i,j)这个点上的结果,即f[i][j]=max(f[i-1][j],f[i][j-1])+a[i][j]

 #include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 110
int a[N][N],f[N][N],n,m;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(f,,sizeof(f));
scanf("%d%d",&n,&m);
for(int i=;i<=n;i++)
{
for(int j=;j<=m;j++)
{
scanf("%d",&a[i][j]);
}
}
for(int i=;i<=n;i++)
{
for(int j=;j<=m;j++)
{
if(i==)f[i][j]=f[i][j-]+a[i][j];
else if(j==)f[i][j]=f[i-][j]+a[i][j];
else f[i][j]=max(f[i-][j],f[i][j-])+a[i][j];
}
}
printf("%d\n",f[n][m]);
}
return ;
}