Little Petya is now fond of data compression algorithms. He has already studied gz, bz, zipalgorithms 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, jthen 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.
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 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.
2 2
1 2
3 4
1 2
2 3
4 3
20 10 30
50 40 30
50 60 70
90 80 70
2 1 3
5 4 3
5 6 7
9 8 7
In the first sample test, despite the fact a1, 2 ≠ a21, they are not located in the same row or column so they may become equal
after the compression.
给一个n*m的矩阵，然后让你压缩一下，输出另外一个n*m的矩阵。
这两个矩阵要求在每一行每一列的大小关系保持不变。比如ai,j<ai,k那么第二个矩阵也得满足这个条件。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long Ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-10;
const int inf =0x7f7f7f7f;
const double pi=acos(-1);
const int maxn=1000000;
int x[maxn+10],y[maxn+10],f[maxn+10],tmp[maxn+10];
int ansx[maxn+10],ansy[maxn+10];
struct node{
int r,l,val,id;
}ne[maxn+10];
bool cmp(node a,node b)
{
return a.val<b.val;
}
bool cmpid(node a,node b)
{
return a.id<b.id;
}
int Find(int i)
{
if(i!=f[i])
f[i]=Find(f[i]);
return f[i];
}
int main()
{
int n,m;
while(~scanf("%d %d",&n,&m))
{
int cnt=0;
MM(tmp,0);
MM(x,0);MM(y,0);
MM(ansx,0);MM(ansy,0);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
scanf("%d",&ne[++cnt].val);
ne[cnt].r=i;
ne[cnt].l=j;
ne[cnt].id=cnt;
f[cnt]=cnt;
}
sort(ne+1,ne+cnt+1,cmp);
for(int k=1;k<=cnt;)
{
int s=k;
for(;ne[k].val==ne[s].val&&k<=cnt;k++);//数值相同的点
for(int i=s;i<k;i++)
{
int r=ne[i].r;
int l=ne[i].l;
if(!x[r]) x[r]=i;
else {
int rr=Find(i);
f[rr]=Find(x[r]);
}
if(!y[l]) y[l]=i;
else {
int rr=Find(i);
f[rr]=Find(y[l]);
}
}
for(int i=s;i<k;i++)
{
int q=Find(i);
int r=ne[i].r,l=ne[i].l;
tmp[q]=max(tmp[q],max(ansx[r],ansy[l])+1);//至少要达到的数值
x[r]=y[l]=0;//清除，最终x,y全都为0，不影响后面
}//核心代码，因为数值从小到大遍历
for(int i=s;i<k;i++)
{
int q=Find(i);
int r=ne[i].r,l=ne[i].l;
ansx[r]=tmp[q];
ansy[l]=tmp[q];
}
}
for(int i=1;i<=cnt;i++)
{
int q=Find(i);
ne[i].val=tmp[q];
}
sort(ne+1,ne+cnt+1,cmpid);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
printf("%d ",ne[(i-1)*m+j].val);
printf("
");
}
}
return 0;
}
分析：很有技术含量的一道题
1：首先将所有的数值从小到大排序，然后开始遍历，每次找出数值相同的点进行跟新，然后开始进行行与列的更新，不过因为同一行或列的点最后压缩出来的数值也得相同，所以需要利用并查集来合并，一同更新；