APIO试题训练

APIO试题训练

---

BZOJ 1177：Oil

（果然我只会做套路题）

考虑3个块的位置，主要可以分为六种情况：

于是我们统计一下左上，右上，左下，右下的最大K*K的矩阵的最大值前缀和，再计算一下以每一行，每一列结尾的最大矩阵，直接枚举切割位置就可以了。

代码：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>

#define rep(i,x,y) for (int i=x;i<=y;i++)
#define dep(i,y,x) for (int i=y;i>=x;i--)
#define sz(x) (int)(x.size())

using namespace std;

typedef long long LL; 
typedef unsigned long long uLL;

const int maxn=1500+7;

int n,m,K,ans,cur;
int a[maxn][maxn],s[maxn][maxn],lu[maxn][maxn],ru[maxn][maxn],ld[maxn][maxn],rd[maxn][maxn],col[maxn],row[maxn];

int calc(int x1,int y1,int x2,int y2)
{
 return s[x2][y2]-s[x1-1][y2]-s[x2][y1-1]+s[x1-1][y1-1];
}

void init()
{
 rep(i,1,n)
 {
  s[i][0]=0;
  rep(j,1,m) s[i][j]=s[i][j-1]+a[i][j];
 }

 rep(i,1,n) rep(j,1,m) s[i][j]+=s[i-1][j];

 rep(i,K,n) rep(j,K,m) lu[i][j]=max(calc(i-K+1,j-K+1,i,j),max(lu[i-1][j],lu[i][j-1]));
 rep(i,K,n) dep(j,m-K+1,1) ru[i][j]=max(calc(i-K+1,j,i,j+K-1),max(ru[i-1][j],ru[i][j+1]));
 dep(i,n-K+1,1) rep(j,K,m) ld[i][j]=max(calc(i,j-K+1,i+K-1,j),max(ld[i+1][j],ld[i][j-1]));
 dep(i,n-K+1,1) dep(j,m-K+1,1) rd[i][j]=max(calc(i,j,i+K-1,j+K-1),max(rd[i+1][j],rd[i][j+1]));

 rep(i,K,n)
 {
  row[i]=0;
  rep(j,K,m) row[i]=max(row[i],calc(i-K+1,j-K+1,i,j));
 }

 rep(j,K,m)
 {
  col[j]=0;
  rep(i,K,m) col[j]=max(col[j],calc(i-K+1,j-K+1,i,j));
 }
}

int main()
{
 scanf("%d%d%d",&n,&m,&K);
 rep(i,1,n) rep(j,1,m) scanf("%d",&a[i][j]);

 init();

 ans=0;
 rep(i,K,n-K)
  rep(j,K,m-K)
  {
   ans=max(ans,lu[i][j]+ru[i][j+1]+rd[i+1][1]);
   ans=max(ans,lu[i][m]+ld[i+1][j]+rd[i+1][j+1]);
  }
 
 rep(j,K,m-K)
  rep(i,K,n-K)
  {
   ans=max(ans,lu[i][j]+ld[i+1][j]+ru[n][j+1]);
   ans=max(ans,lu[n][j]+ru[i][j+1]+rd[i+1][j+1]);
  }
 
 rep(i,K,n-2*K)
 {
  cur=0;
  rep(j,i+K,n-K)
  {
   cur=max(cur,row[j]);
   ans=max(ans,lu[i][m]+cur+ld[j+1][m]);
  }
 }
 
 rep(i,K,m-2*K)
 {
  cur=0;
  rep(j,i+K,m-K)
  {
   cur=max(cur,col[j]);
   ans=max(ans,lu[n][i]+cur+ru[n][j+1]);
  }
 }

 printf("%d\n",ans);

 // system("pause");
 return 0;
}