題目大意:
思路;
s[i][j]=s[i−1][j]+s[i][j−1]−s[i−1][j−1]+a[i][j]s[i][j]=s[i−1][j]+s[i][j−1]−s[i−1][j−1]+a[i][j]
那麼再枚舉正方形的右下角求出每個矩陣的答案,再取最大值即可。
ans=max(s[i][j]−s[i−m][j]−s[i][j−m]+s[i−m][j−m])ans=max(s[i][j]−s[i−m][j]−s[i][j−m]+s[i−m][j−m])
代碼:
#include <cstdio>
#include <iostream>
using namespace std;
int n,m,s[5011][5011],x,y,lenn,lenm,ans;
int main()
{
scanf("%d%d",&n,&m);
for (int i=1;i<=n;i++)
{
scanf("%d%d",&x,&y);
scanf("%d",&s[x+1][y+1]);
}
for (int i=1;i<=5001;i++)
for (int j=1;j<=5001;j++)
s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+s[i][j]; //二維字首和
for (int i=m;i<=5001;i++)
for (int j=m;j<=5001;j++)
ans=max(ans,s[i][j]-s[i-m][j]-s[i][j-m]+s[i-m][j-m]);
printf("%d\n",ans);
return 0;
}