POJ2019: Two-dimensional St algorithm for solving the problem of the maximum value of static matrices

We really need to extend the ST algorithm to two-dimensional, because the two-dimensional rmq problem is still a lot of

int n,b,k; int mm[505]; int VAL[MAXN][MAXN]; int dpmin[maxn][maxn][8[8]; int dpmax[maxn][maxn][8[8];

Here n is the square of the length of the square, here is the topic, and then mm is preprocessed, convenient to calculate the index

Dpmin and Dpmax are preprocessing arrays.

Then take a look at the start pretreatment:

voidINITRMQ (intNintm) {     for(intI=1; i<=n;i++)     for(intj=1; j<=m;j++) dpmin[i][j][0][0]=dpmax[i][j][0][0]=Val[i][j];  for(intIi=0; ii<=mm[n];ii++)     for(intjj=0; jj<=mm[m];jj++)    if(ii+JJ) for(intI=1; i+ (1&LT;&LT;II)-1<=n;i++)     for(intj=1; j+ (1&LT;&LT;JJ)-1<=m;j++)    {        if(ii) {DPMIN[I][J][II][JJ]= Min (dpmin[i][j][ii-1][jj],dpmin[i+ (1<< (ii-1))][j][ii-1][JJ]); DPMAX[I][J][II][JJ]= Max (dpmax[i][j][ii-1][jj],dpmax[i+ (1<< (ii-1))][j][ii-1][JJ]); }        Else{DPMIN[I][J][II][JJ]= Min (dpmin[i][j][ii][jj-1],dpmin[i][j+ (1<< (jj-1))][ii][jj-1]); DPMAX[I][J][II][JJ]= Max (dpmax[i][j][ii][jj-1],dpmax[i][j+ (1<< (jj-1))][ii][jj-1]); }    }}

We look at the pre-processing time is still relatively clear, of course, do not forget the main function of the MM initialization good

    mm[0]=-1;      for (int i=1; i<=; i++)        mm[i]= ((i& (-1)) = =0) mm[i-1]+1: mm[i-1];

Then it is the function that asks for the maximum and minimum value, here, must be careful to write, it is easy to write wrong:

intRMQ1 (intX1,intY1,intX2,inty2)//Max{    intk1=mm[x2-x1+1]; intk2=mm[y2-y1+1]; X2=x2-(1<<K1) +1; Y2=y2-(1<<K2) +1; returnMax ( Max (Dpmax[x1][y1][k1][k2],dpmax[x1][y2][k1][k2]), Max (DPMAX[X2][Y1][K1][K2],DPMAX[X2][Y2][K1][K2]));}intRMQ2 (intX1,intY1,intX2,inty2) {    intk1=mm[x2-x1+1]; intk2=mm[y2-y1+1]; X2=x2-(1<<K1) +1; Y2=y2-(1<<K2) +1; returnmin (min (dpmin[x1][y1][k1][k2],dpmin[x1][y2][k1][k2]), Min (Dpmin[x2][y1][k1][k2],dpmin[x2][y2][k1][k2]));}

That's a really long one.

Finally, the complete realization of the topic is given:

1#include <cstdio>2#include <algorithm>3 using namespacestd;4 Const intmaxn=255;5 intn,b,k;6 intmm[505];7 intVAL[MAXN][MAXN];8 intdpmin[maxn][maxn][8][8];9 intdpmax[maxn][maxn][8][8];Ten voidINITRMQ (intNintm) One { A      for(intI=1; i<=n;i++) -      for(intj=1; j<=m;j++) -dpmin[i][j][0][0]=dpmax[i][j][0][0]=Val[i][j]; the      for(intIi=0; ii<=mm[n];ii++) -      for(intjj=0; jj<=mm[m];jj++) -     if(ii+JJ) -      for(intI=1; i+ (1&LT;&LT;II)-1<=n;i++) +      for(intj=1; j+ (1&LT;&LT;JJ)-1<=m;j++) -     { +         if(ii) A         { atDPMIN[I][J][II][JJ] = min (dpmin[i][j][ii-1][jj],dpmin[i+ (1<< (ii-1))][j][ii-1][JJ]); -DPMAX[I][J][II][JJ] = max (dpmax[i][j][ii-1][jj],dpmax[i+ (1<< (ii-1))][j][ii-1][JJ]); -         } -         Else -         { -DPMIN[I][J][II][JJ] = min (dpmin[i][j][ii][jj-1],dpmin[i][j+ (1<< (jj-1))][ii][jj-1]); inDPMAX[I][J][II][JJ] = max (dpmax[i][j][ii][jj-1],dpmax[i][j+ (1<< (jj-1))][ii][jj-1]); -         } to     } + } - intRMQ1 (intX1,intY1,intX2,inty2)//Max the { *     intk1=mm[x2-x1+1]; $     intk2=mm[y2-y1+1];Panax Notoginsengx2=x2-(1&LT;&LT;K1) +1; -y2=y2-(1&LT;&LT;K2) +1; the     returnMax ( Max (DPMAX[X1][Y1][K1][K2],DPMAX[X1][Y2][K1][K2), Max (DPMAX[X2][Y1][K1][K2],DPMAX[X2][Y2][K1][K2])); + } A intRMQ2 (intX1,intY1,intX2,inty2) the { +     intk1=mm[x2-x1+1]; -     intk2=mm[y2-y1+1]; $x2=x2-(1&LT;&LT;K1) +1; $y2=y2-(1&LT;&LT;K2) +1; -     returnmin (min (dpmin[x1][y1][k1][k2],dpmin[x1][y2][k1][k2]), Min (Dpmin[x2][y1][k1][k2],dpmin[x2][y2][k1][k2])); - } the intMain () - {Wuyimm[0]=-1; the      for(intI=1; i<= -; i++) -Mm[i]= ((i& (i-1))==0)? mm[i-1]+1: mm[i-1]; Wu      while(SCANF ("%d%d%d", &n,&b,&k) = =3) -     { About          for(intI=1; i<=n;i++) $          for(intj=1; j<=n;j++) -scanf"%d",&val[i][j]); - INITRMQ (n,n); -         intx, y; A          while(k--) +         { thescanf"%d%d",&x,&y); -printf"%d\n", RMQ1 (x,y,x+b-1, y+b-1)-RMQ2 (x,y,x+b-1, y+b-1)); $         } the     } the     return 0; the}

POJ2019: Two-dimensional St algorithm for solving the problem of the maximum value of static matrices