Code Forces 650 C Table Compression(并查集)

C. Table Compression
time limit per test4 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zip algorithms and many others. Inspired by the new knowledge, Petya is now developing the new compression algorithm which he wants to name dis.

Petya decided to compress tables. He is given a table a consisting of n rows and m columns that is filled with positive integers. He wants to build the table a’ consisting of positive integers such that the relative order of the elements in each row and each column remains the same. That is, if in some row i of the initial table ai, j < ai, k, then in the resulting table a’i, j < a’i, k, and if ai, j = ai, k then a’i, j = a’i, k. Similarly, if in some column j of the initial table ai, j < ap, j then in compressed table a’i, j < a’p, j and if ai, j = ap, j then a’i, j = a’p, j.

Because large values require more space to store them, the maximum value in a’ should be as small as possible.

Petya is good in theory, however, he needs your help to implement the algorithm.

Input
The first line of the input contains two integers n and m (, the number of rows and the number of columns of the table respectively.

Each of the following n rows contain m integers ai, j (1 ≤ ai, j ≤ 109) that are the values in the table.

Output
Output the compressed table in form of n lines each containing m integers.

If there exist several answers such that the maximum number in the compressed table is minimum possible, you are allowed to output any of them.

Examples
input
2 2
1 2
3 4
output
1 2
2 3
input
4 3
20 10 30
50 40 30
50 60 70
90 80 70
output
2 1 3
5 4 3
5 6 7
9 8 7

把一个二维数组离散化
先把数组里的数全部排序。用并查集把同一行或者同一列的相同元素并成一个集合，这个集合里的每个数的离散化后的序号一定是相同的，因为他们一样大并且在同一行或者同一列。每个元素肯定是其所在行和所在列的最大值加1，那么选取这些元素的中最大的作为根的值，其余的元素的值都和根的值相等，具体见代码

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <string>

using namespace std;
#define MAX 1000000
struct Node
{
   int x;int y;
   int value;
}b[MAX+5];
int n,m;
int cmp(Node a,Node b)
{
    return a.value<b.value;
}
int father[MAX+5];
int r[MAX+5];
int l[MAX+5];
int mr[MAX+5];
int ml[MAX+5];
int ans[MAX+5];
int get(int x)
{
    if(x!=father[x])
        father[x]=get(father[x]);
    return father[x];
}
void join(int x,int y)
{
    int fx=get(x);
    int fy=get(y);
    if(fx!=fy)
        father[fx]=fy;
}
int main()
{
    scanf("%d%d",&n,&m);
    int cnt=1;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
           scanf("%d",&b[(i-1)*m+j].value);
              b[cnt].x=i;b[cnt++].y=j;
        }
    }
    sort(b+1,b+n*m+1,cmp);
    for(int i=1;i<=n*m;i++)
        father[i]=i;
   int i=1,k;
   while(i<=n*m)
   {
        for( k=i;k<=n*m;k++)
        {
            if(b[k].value!=b[i].value)
                break;
        }
        for(int j=i;j<k;j++)
        {
            if(!r[b[j].x])
                r[b[j].x]=(b[j].x-1)*m+b[j].y;
            else
            {
                join(r[b[j].x],(b[j].x-1)*m+b[j].y);
            }
            if(!l[b[j].y])
                l[b[j].y]=(b[j].x-1)*m+b[j].y;
            else
            {
                join(l[b[j].y],(b[j].x-1)*m+b[j].y);
            }
        }
        for(int j=i;j<k;j++)
        {
            int s=get((b[j].x-1)*m+b[j].y);
            ans[s]=max(ans[s],max(mr[b[j].x],ml[b[j].y])+1);
        }
        for(int j=i;j<k;j++)
        {
            int s=get((b[j].x-1)*m+b[j].y);
            ans[(b[j].x-1)*m+b[j].y]=ans[s];
            mr[b[j].x]=ans[s];ml[b[j].y]=ans[s];
            r[b[j].x]=0;l[b[j].y]=0;
        }
        i=k;
    }
    for(int i=1;i<=n*m;i++)
    {
        if((i)%m==0)
            printf("%d
",ans[i]);
        else
            printf("%d ",ans[i]);
    }
    return 0;
}