Codeforces Round #345 (Div. 1) C. Table Compression dp+并查集

题目链接：

http://codeforces.com/problemset/problem/650/C

C. Table Compression

time limit per test4 seconds
memory limit per test256 megabytes #### 问题描述 > 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. #### 输入 > 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. #### 输出 > 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. ####样例输入 > 2 2 > 1 2 > 3 4

样例输出

1 2
2 3

题意

给你一个n*m的方格，每个方格上有一个正整数，现在要把这些数离散化，要保证离散化之后同一行或同一列的数之间大小关系不变，构造一种解使得最大的那个值最小

题解

[Portal]

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef __int64 LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=1e6+10;

int n,m;
int val[maxn],dp[maxn],ra[maxn],ind[maxn];
VI G[maxn];
VPII egs;

int fa[maxn];
int find(int x){
    return fa[x]=fa[x]==x?x:find(fa[x]);
}

bool cmp(int a,int b){
    return val[a]<val[b];
}

void init(){
    for(int i=0;i<=n*m;i++) fa[i]=i,dp[i]=-1,ind[i]=0;
}

int main() {
    scf("%d%d",&n,&m);
    init();
    rep(i,0,n) rep(j,0,m){
        scf("%d",&val[i*m+j]);
    }

    //每行/每列 排序,然后连成串(这样可以少连很多没有用的边
    rep(i,0,n){
        rep(j,0,m) ra[j]=i*m+j;
        sort(ra,ra+m,cmp);
        for(int j=0;j<m-1;j++){
            int p1=ra[j],p2=ra[j+1];
            if(val[p1]<val[p2]){
                egs.pb(mkp(p1,p2));
            }else{
                fa[find(p2)]=fa[find(p1)];
            }
        }
    }

    rep(j,0,m){
        rep(i,0,n) ra[i]=i*m+j;
        sort(ra,ra+n,cmp);
        for(int i=0;i<n-1;i++){
            int p1=ra[i],p2=ra[i+1];
            if(val[p1]<val[p2]){
                egs.pb(mkp(p1,p2));
            }else{
                fa[find(p2)]=fa[find(p1)];
            }
        }
    }

    //用并查集缩点
    for(int i=0;i<egs.sz();i++){
        int pu=find(egs[i].X),pv=find(egs[i].Y);
        G[pu].pb(pv);
        ind[pv]++;
    }

    //拓扑排序
    queue<int> Q;

    rep(i,0,n*m){
        if(fa[i]!=i) continue;
        if(ind[i]==0){
            dp[i]=1;
            Q.push(i);
        }
    }

    while(!Q.empty()){
        int u=Q.front(); Q.pop();
        for(int i=0;i<G[u].sz();i++){
            int v=G[u][i];
            dp[v]=max(dp[v],dp[u]+1);
            ind[v]--;
            if(ind[v]==0) Q.push(v);
        }
    }

    rep(i,0,n){
        rep(j,0,m){
            int x=dp[find(i*m+j)];
            prf("%d",x);
            if(j==m-1) puts("");
            else prf(" ");
        }
    }

    return 0;
}

//end-----------------------------------------------------------------------