[2017紀中10-24]方陣 二維ST表
阿新 • • 發佈：2019-01-26
題面
假的資料。。。考場上被坑成0分。
首先兩個log的二維ST表很好想，st[i][j][k][l]表示以點（i，j）向右2^k，向下2^l的矩形內的答案。
（假設）考慮真的長不超過寬的兩倍，我們可以把它分成兩個以寬為邊長的正方形（當然可能會有部分重疊），那麼詢問都轉化成正方形，只需要預處理st[i][j][k]表示以點（i，j）向右下2^k的正方形內的答案。時間和空間都少了一個log。
原版程式碼：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath> 但因為資料的原因，要把它分成多個正方形，於是寫一個迴圈來分。
迴圈版程式碼：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define chkmin(a,b) a=min(a,b)
#define chkmax(a,b) a=max(a,b)
using namespace std;
int n,m,w,q,a[810][810],mx[810][810][11],mi[810][810][11];
long long sum[810][810];
int qmx(int x1,int y1,int x2,int y2,int k)
{
int t=(int)(log(k)/log(2)),r=(1<<t);
return max(max(mx[x1][y1][t],mx[x2-r+1][y2-r+1][t]),max(mx[x2-r+1][y1][t],mx[x1][y2-r+1][t]));
}
int qmi(int x1,int y1,int x2,int y2,int k)
{
int t=(int)(log(k)/log(2)),r=(1<<t);
return min(min(mi[x1][y1][t],mi[x2-r+1][y2-r+1][t]),min(mi[x2-r+1][y1][t],mi[x1][y2-r+1][t]));
}
int spmx(int x1,int y1,int x2,int y2,int k)
{
int re=0;
if(x1+k==x2)
{
for(int t=1;y1+k<y2;t++,y1+=k+1) chkmax(re,qmx(x1,y1,x2,y1+k,k+1));
chkmax(re,qmx(x1,y2-k,x2,y2,k+1));
}
else
{
for(int t=1;x1+k<x2;t++,x1+=k+1) chkmax(re,qmx(x1,y1,x1+k,y2,k+1));
chkmax(re,qmx(x2-k,y1,x2,y2,k+1));
}
return re;
}
int spmi(int x1,int y1,int x2,int y2,int k)
{
int re=1e9;
if(x1+k==x2)
{
for(int t=1;y1+k<y2;t++,y1+=k+1) chkmin(re,qmi(x1,y1,x2,y1+k,k+1));
chkmin(re,qmi(x1,y2-k,x2,y2,k+1));
}
else
{
for(int t=1;x1+k<x2;t++,x1+=k+1) chkmin(re,qmi(x1,y1,x1+k,y2,k+1));
chkmin(re,qmi(x2-k,y1,x2,y2,k+1));
}
return re;
}
int main()
{
freopen("phalanx.in","r",stdin);
freopen("phalanx.out","w",stdout);
scanf("%d%d",&n,&m);
w=min(n,m);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
scanf("%d",&a[i][j]);
memset(mi,0x3f,sizeof(mi));
memset(mx,0,sizeof(mx));
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+a[i][j];
mx[i][j][0]=mi[i][j][0]=a[i][j];
}
for(int k=1;(1<<k)<=w;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
mx[i][j][k]=mx[i][j][k-1];
if(i+(1<<(k-1))<=n) chkmax(mx[i][j][k],mx[i+(1<<(k-1))][j][k-1]);
if(j+(1<<(k-1))<=m) chkmax(mx[i][j][k],mx[i][j+(1<<(k-1))][k-1]);
if(i+(1<<(k-1))<=n&&j+(1<<(k-1))<=m) chkmax(mx[i][j][k],mx[i+(1<<(k-1))][j+(1<<(k-1))][k-1]);
}
for(int k=1;(1<<k)<=w;k++)
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
mi[i][j][k]=mi[i][j][k-1];
if(i+(1<<(k-1))<=n) chkmin(mi[i][j][k],mi[i+(1<<(k-1))][j][k-1]);
if(j+(1<<(k-1))<=m) chkmin(mi[i][j][k],mi[i][j+(1<<(k-1))][k-1]);
if(i+(1<<(k-1))<=n&&j+(1<<(k-1))<=m) chkmin(mi[i][j][k],mi[i+(1<<(k-1))][j+(1<<(k-1))][k-1]);
}
scanf("%d",&q);
for(int i=1;i<=q;i++)
{
char opt[3];
int x1,y1,x2,y2,k;
scanf("%s",opt);
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);x1++;y1++;x2++;y2++;
k=min(x2-x1,y2-y1);
if(opt[1]=='U') {printf("%lld\n",sum[x2][y2]-sum[x1-1][y2]-sum[x2][y1-1]+sum[x1-1][y1-1]);continue;}
if(opt[1]=='A') {printf("%d\n",spmx(x1,y1,x2,y2,k));}
if(opt[1]=='I') {printf("%d\n",spmi(x1,y1,x2,y2,k));}
}
return 0;
}