HDU 4941 magical forest [discretization] [map]

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4941

Here are 10 ^ 5 vertices. Each vertex has a numerical value. The XY coordinate of the vertex is 0 ~ 10 ^ 9, points exist in the matrix. Then, a 10 ^ 5 operation is given. 1 indicates the exchange row, 2 indicates the exchange column, and 3 indicates the value where the query coordinate is the XY point.

A large amount of data...

After the competition, I saw that the problem was solved by first discretization and then linear map.

#include <iostream>#include <cstdio>#include <map>#include <algorithm>using namespace std;struct Node{    int x;    int y;    int c;}l[100000];map <int,  int>  fmap[100010];map <int ,int>hashx,hashy;int T,n,m,k,mapx,mapy,t,cas,lianx[100010],liany[100010],Q,A,B,X,Y;bool cmpx(Node a,Node b){    return a.x<b.x;}bool cmpy(Node a,Node b){    return a.y<b.y;}int main(){    int CAS;    scanf("%d", &CAS);    for(int cas=1;cas<=CAS;cas++)    {      scanf("%d%d%d",&n,&m,&k);      printf("Case #%d:\n",cas);      for(int i=0;i<k;i++) scanf("%d%d%d",&l[i].x,&l[i].y,&l[i].c);      for(int i=0;i<100010;i++) fmap[i].clear();        hashx.clear();  hashy.clear();  mapx=mapy=0;      sort(l,l+k,cmpx);      for(int i=0;i<k;i++) if(!hashx[l[i].x]) hashx[l[i].x]=++mapx;      sort(l,l+k,cmpy);      for(int i=0;i<k;i++)      {        if(!hashy[l[i].y]) hashy[l[i].y]=++mapy;        fmap[hashx[l[i].x]][hashy[l[i].y]]=l[i].c;      }      for(int i=1;i<=mapx;i++)lianx[i]=i;      for(int i=1;i<=mapy;i++)liany[i]=i;      scanf("%d",&t);      for(int i=0;i<t;i++)      {        scanf("%d%d%d",&Q,&A,&B);        if(Q==1)        {          X=hashx[A];Y=hashx[B];          if(X && Y) swap(lianx[X],lianx[Y]);        }        else if(Q==2)        {          X=hashy[A];Y=hashy[B];          if(X && Y) swap(liany[X],liany[Y]);        }        else if(Q==3)        {          X=hashx[A];Y=hashy[B];          if(X && Y)printf("%d\n",fmap[lianx[X]][liany[Y]]);          else printf("0\n");        }      }    }    return 0;}