Coj 2105 Submatrix

Submatrix

Difficulty level: A; programming language: Unlimited; run time limit: 2000ms; run space limit: 131072KB; code length limit: 102400B

Question Description

Small A has a nxm matrix in which 1~n*m (n*m) integers appear once. Now small a selects a sub-matrix within the matrix whose weights are equal to the minimum value of all the numbers in the sub-matrix. Little A would like to know if the weight of the sub-matrix he chose was I (1<=I<=NXM), then how many of the sub-matrices he chose might be? Little a wants to know the result of all possible I values, but these results are too many, he can't count, so he asks you for help.

Input

First line, two integers n, M. The next n rows, m integers per line, represent the elements in the matrix.

Output

NXM lines, one integer per line, where the integer of line I represents the number of sub-matrices that he chooses if the sub-matrix weights for small A are selected as I.

Input example

2 3 2 5 1 6 3 4

Output example

6 4 5 1 1 1

Other Notes

For 30% of data, 1<=n, m<=50; For all the data, 1<=n, m<=300.

Solution:%%% Zheng, first consider a one-dimensional, with a monotonous stack maintenance can go to the left to the right, update the answer. For two-dimensional, the one-dimensional beat can be .... Remember to take the minimum value ...

1#include <iostream>2#include <cstdio>3#include <cmath>4#include <algorithm>5#include <queue>6#include <cstring>7 #definePAU Putchar (")8 #defineENT Putchar (' \ n ')9 using namespacestd;Ten Const intmaxn= -+Ten, maxans=90000+Ten, inf=-1u>>1; One inta[maxn][maxn],mi[maxn],r[maxn],l[maxn],s[maxn],pos[maxn],p[maxans],n,m; AInlineintRead () { -     intx=0, sig=1;CharCh=GetChar (); -      for(;! IsDigit (CH); Ch=getchar ())if(ch=='-') sig=0; the      for(; isdigit (ch); Ch=getchar ()) x=Ten*x+ch-'0'; -     returnsig?x:-x; - } -InlinevoidWriteintx) { +     if(x==0) {Putchar ('0');return;}if(x<0) Putchar ('-'), x=-x; -     intlen=0, buf[ the]; while(x) buf[len++]=x%Ten, x/=Ten; +      for(inti=len-1; i>=0; i--) Putchar (buf[i]+'0');return; A } at voidinit () { -N=read (); m=read (); -      for(intI=1; i<=n;i++) for(intj=1; j<=m;j++) a[i][j]=read (); -      for(intI=1; i<=n;i++){ -          for(intj=1; j<=m;j++) mi[j]=inf; -          for(intj=i;j<=n;j++){ in              for(intk=1; k<=m;k++) Mi[k]=min (Mi[k],a[j][k]);inttop=0; -              for(intk=1; k<=m;k++) {//is to                  while(top&& (Mi[k]<s[top])) r[pos[top--]]=k-1; +s[++top]=mi[pos[top]=K]; -} while(top) r[pos[top--]]=m; the              for(intk=m;k>=1; k--) {//Anti- *                  while(top&& (Mi[k]<s[top])) l[pos[top--]]=k+1; $s[++top]=mi[pos[top]=K];Panax Notoginseng} while(top) l[pos[top--]]=1; -              for(intk=1; k<=m;k++) p[mi[k]]+= (r[k]-k+1) * (k-l[k]+1); the         } +}intAll=n*m; for(intI=1; i<=all;i++) write (P[i]), ENT; A     return; the } + voidWork () { -     return; $ } $ voidprint () { -     return; - } the intMain () {init (); work ();p rint ();return 0;}

Coj 2105 Submatrix