找出三个正方形，可以转化为将整个油田切成三个矩形块，每块中各找一个正方形区域，切的形式只有6种，分类更新ans即可
题解：http://trinklee.blog.163.com/blog/static/238158060201482371229105/
另：这题一般的快速读入不知为何会RE，但是题解里这位大神的快速读入就能AC……跪了跪了
1 /************************************************************** 2 Problem: 1177 3 User: Tunix 4 Language: C++ 5 Result: Accepted 6 Time:6088 ms 7 Memory:62028 kb 8 ****************************************************************/ 9 10 //BZOJ 1177 11 #include<vector> 12 #include<cstdio> 13 #include<cstring> 14 #include<cstdlib> 15 #include<iostream> 16 #include<algorithm> 17 #define rep(i,n) for(int i=0;i<n;++i) 18 #define F(i,j,n) for(int i=j;i<=n;++i) 19 #define D(i,j,n) for(int i=j;i>=n;--i) 20 #define pb push_back 21 using namespace std; 22 inline int getint(){ 23 int v=0,sign=1; char ch=getchar(); 24 while(ch<'0'||ch>'9'){ if (ch=='-') sign=-1; ch=getchar();} 25 while(ch>='0'&&ch<='9'){ v=v*10+ch-'0'; ch=getchar();} 26 return v*sign; 27 } 28 const int N=1610,INF=~0u>>2; 29 typedef long long LL; 30 /******************tamplate*********************/ 31 int n,m,k; 32 typedef int Matrix[N][N]; 33 Matrix a,b,c,d,s,t; 34 int ans; 35 36 int main(){ 37 #ifndef ONLINE_JUDGE 38 freopen("1177.in","r",stdin); 39 freopen("1177.out","w",stdout); 40 #endif 41 n=getint(); m=getint(); k=getint(); 42 int x; 43 F(i,1,n) F(j,1,m){ 44 scanf("%d",&x); 45 t[i][j]=t[i-1][j]+t[i][j-1]-t[i-1][j-1]+x; 46 } 47 F(i,k,n) F(j,k,m) s[i][j]=t[i][j]-t[i-k][j]-t[i][j-k]+t[i-k][j-k]; 48 49 F(i,k,n) F(j,k,m) a[i][j]=max(s[i][j],max(a[i-1][j],a[i][j-1])); 50 F(i,k,n) D(j,m,k) b[i][j]=max(s[i][j],max(b[i-1][j],b[i][j+1])); 51 D(i,n,k) F(j,k,m) c[i][j]=max(s[i][j],max(c[i+1][j],c[i][j-1])); 52 D(i,n,k) D(j,m,k) d[i][j]=max(s[i][j],max(d[i+1][j],d[i][j+1])); 53 54 F(i,k,n-k) F(j,k,m-k) ans=max(ans,a[i][j]+b[i][j+k]+c[i+k][m]); 55 F(i,k,n-k) F(j,k,m-k) ans=max(ans,a[i][j]+c[i+k][j]+b[n][j+k]); 56 F(i,k,n-k) F(j,k,m-k) ans=max(ans,a[i][m]+c[i+k][j]+d[i+k][j+k]); 57 F(i,k,n-k) F(j,k,m-k) ans=max(ans,a[n][j]+b[i][j+k]+d[i+k][j+k]); 58 F(i,k,n) F(j,k+k,m-k) ans=max(ans,s[i][j]+a[n][j-k]+b[n][j+k]); 59 F(i,k+k,n-k) F(j,k,m) ans=max(ans,s[i][j]+a[i-k][m]+c[i+k][m]); 60 printf("%d\n",ans); 61 return 0; 62 }