Hdu1338 learned and Heroes College

Address: http://acm.uestc.edu.cn/#/problem/show/1338

Ideas:

learned and Heroes CollegeTime limit:6000/2000ms (java/others) Memory limit:225535/225535kb (java/others)SubmitStatus

Most humans, in this era, have the power to be called "personalities," but those who have power do not necessarily belong to the righteous. Wherever evil arises, there must be a hero who will come forward to save the crowd. A young man born without power------learned dreamed of a top hero since childhood, and his dream is to become a great hero, but he can realize his dream without power? Despite the difficulties, the teenager still does not give up, toward his own goal to go forward bravely!

......

Although learned has no personality, learned is indeed a "hero" with a super-strong will.

The so-called heroes are able to stand up when everyone is indifferent!

Today, learned appeared!

Learned faces a matrix of n∗mn∗m, each of which has a monster in each position.

Learned in order to better defeat them, his first task is to analyze the combat effectiveness of monsters.

In order to simplify the calculation of your analysis time, learned need to get another n∗mn∗m matrix.

This matrix satisfies the following requirements:

1. In each row of this new matrix, the size relationship in each column is the same as the original matrix.

2. The minimum value of this new matrix should be as small as possible, so that the smallest value is minimized, so that the second small value is as small as possible ... So that the maximum value is as small as possible.

Now learned is very busy, please help learned output this matrix!

Input

The first line of N,m represents the size of the matrix

Next n lines, m integers per line, represents the combat effectiveness of monsters

Guarantee:

1<=n∗m<=1000000 1<=n∗m<=1000000

The monster's combat effectiveness is the number within the INT range.

Output

To output a new matrix ~

The smallest number in the new matrix is at least 1 oh.

Sample Input and output

Sample Input	Sample Output
3 31 2 34 5 67 8 9	1 2 32 3 43 4 5
2 21 43 2	1 22 1

Ideas:

The most basic idea is to order all the numbers, and then fill in one, but because the same number of peers in the same column is troublesome.

So every time you drag all the same number of peers in the same column into the same set, and check the set to maintain the same number of values should be put

Just such a layer of consolidation, until filled out.

But the output of the time to use and check the set of the answer recorded in the output ... (saying the question has been thinking about me for 10,000 years,,,,

1#include <iostream>2#include <algorithm>3#include <cstdio>4#include <cmath>5#include <cstring>6#include <queue>7#include <stack>8#include <map>9#include <vector>Ten#include <cstdlib> One#include <string> A  - #definePI ACOs ((double)-1) - #defineE exp (double (1)) the Const intmax=1e6+Ten; - using namespacestd; - structnode - { +     intv,x,y; - }a[max]; + intAns[max],xmax[max],ymax[max],hx[max],hy[max]; A intRoot[max]; at intFdintx) - { -     returnx!=root[x]?root[x]=FD (Root[x]): x; - } - voidJoinintXinty) - { in     intA=FD (x), b=fd (y); -     if(a!=b) root[a]=b; to } + BOOLcmpstructNode C,structNode B) - { the     returnc.v<B.V; * } $ intMain (void)Panax Notoginseng { -     intn,m; theCin>>n>>m; +      for(intI=1; i<=n;i++) A          for(intj=1; j<=m;j++) the         { +scanf"%d", &a[(i-1) *m+j].v); -a[(I-1) *m+j].x=i;a[(I-1) *m+j].y=J; $         } $Sort (A +1, a+n*m+1, CMP); -memset (Xmax,0,sizeof(Xmax)); -memset (Ymax,0,sizeof(Ymax));//all set to 0, thememset (ans,0,sizeof(ans)); -     intSame=1;//record the same elementsWuyi      for(intI=1; i<=n*m;i++) theroot[i]=i; -      for(intI=1; i<=n*m;i++) Wu     { -         if(I!=n*m && a[i].v==a[i+1].v) About             Continue; $          for(intj=same;j<=i;j++) -         { -Hx[a[j].x]=hy[a[j].y]= (a[j].x-1) *m+a[j].y; -         } A          for(intj=same;j<=i;j++) +         { theJoin (hx[a[j].x], (a[j].x-1) *m+a[j].y); -Join (Hy[a[j].y], (a[j].x-1) *m+a[j].y); $         } the          for(intj=same;j<=i;j++) the         { the             intx = a[j].x,y=a[j].y; the             intRT = FD ((a[j].x-1) *m+a[j].y); -Ans[rt]=max (Ans[rt],max (xmax[x],ymax[y]) +1); in         } the          for(intj=same;j<=i;j++) the         { About             intx = a[j].x,y=a[j].y; theXmax[x]=max (XMAX[X],ANS[FD (a[j].x-1) *m+a[j].y)]); theYmax[y]=max (YMAX[Y],ANS[FD (a[j].x-1) *m+a[j].y)]); the         } +same=i+1; -     } the      for(intI=1; i<=n*m;i++)Bayi     { the         if(i%m==0) theprintf"%d \ n", ANS[FD (i)]); -         Else -printf"%d", ANS[FD (i)]); the     } the     return 0; the}

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Hdu1338 learned and Heroes College