Codeforces Round #593 (Div. 2) D. Alice and the Doll

Topic: http://codeforces.com/problemset/problem/1236/D
ideas: robots can only follow the law → ↓ ← ↑ to move, so the most distant in the current direction can move forward is the limit, the right to reach the most distant turn, and the next can not then move farther in that direction, and thus according to simulation → ↓ ← ↑, updated every movement limit, then the end can not be moved (first movement can not turn right, n * m will burst int)
          barriers exist with the set, each half looking to reach the farthest place, and compared with the limits

#include<bits/stdc++.h>
 
using namespace std;
 
#define pb push_back
#define fi first
#define se second
#define mem(i,j) memset(i,j,sizeof(i))
 
typedef long long ll;
typedef pair<int,int>pii;
 
const int maxn=1e5+5;
 
ll n,m;
int k;
int a,b;
set<int>x[maxn],y[maxn];
 
int main ()
{
    scanf("%I64d%I64d%d",&n,&m,&k);
    for(int i=1;i<=k;i++)
    {
        scanf("%d%d",&a,&b);
        x[a].insert(b);
        y[b].insert(a);
    } 
    int turn=1;
    int xx=1,yy=1;
    ll cnt=1;
    int t;
    int minx=2,maxx=n,miny=1,maxy=m;
    while(1)
    {
        if(turn==1)
        {
            t=yy;
            auto it=x[xx].lower_bound(yy);
            if(it==x[xx].end()) yy=m;
            else yy=*it-1;
            if(yy>maxy) yy=maxy;
            maxy=yy-1;
            if(t-yy==0&&xx!=1&&yy!=1) break;
            cnt+=yy-t;
        }
        else if(turn==2)
        {
            t=xx;
            auto it=y[yy].lower_bound(xx);
            if(it==y[yy].end()) xx=n;
            else xx=*it-1;
            if(xx>maxx) xx=maxx;
            maxx=xx-1;
            if(t-xx==0) break;
            cnt+=xx-t;
        }
        else if(turn==3)
        {
            t=yy;
            auto it=x[xx].lower_bound(yy);
            if(it==x[xx].begin()) yy=1;
            else yy=*(--it)+1;
            if(yy<miny) yy=miny;
            miny=yy+1;    
            if(t-yy==0) break;
            cnt+=t-yy;
        }
        else if(turn==0)
        {
            t=xx;
            auto it=y[yy].lower_bound(xx);
            if(it==y[yy].begin()) xx=1;
            else xx=*(--it)+1;
            if(xx<minx) xx=minx;
            minx=xx+1;
            if(t-xx==0) break;
            cnt+=t-xx;
        }
        if(cnt==n*m-k) {printf("Yes\n");return 0;}
        turn=(turn+1)%4;
        
    }
    printf("No\n");
    return 0;
}