HDU 6229 Wandering Robots 找规律+离散化

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=3e3+10;
bool vis[N][N],mp[N][N];
int dx[]={0,1,0,-1};
int dy[]={1,0,-1,0};
int n,k,mm,mn;
int sc[N],sr[N],cn1,cn2;
int x[N],y[N];
int x_hash(int x)
{
    return lower_bound(sc,sc+mn,x)-sc;
}
int y_hash(int y)
{
    return lower_bound(sr,sr+mm,y)-sr;
}
void dfs(int x,int y)
{
    vis[x][y]=1;
    for(int i=0;i<4;i++)
    {
        int tx=x+dx[i],ty=y+dy[i];
        if(tx>=0&&tx<mn&&ty>=0&&ty<mm&&!vis[tx][ty]&&!mp[tx][ty])dfs(tx,ty);
    }
}
int get(int x,int y)
{
    if(!vis[x][y])return 0;
    int cnt=vis[x][y];
    for(int i=0;i<4;i++)
    {
       int tx=x+dx[i],ty=y+dy[i];
        if(tx>=0&&tx<mn&&ty>=0&&ty<mm)cnt+=vis[tx][ty];
    }
    return cnt;
}
int main(){
    int T;
    scanf("%d",&T);
    int cas=1;
    while(T--)
    {
        cn1=0;cn2=0;
        //memset(vis,0,sizeof(vis));
        //memset(mp,0,sizeof(mp));
        scanf("%d %d",&n,&k);
        sc[cn1++]=0;sc[cn1++]=n-1;
        sr[cn2++]=0;sr[cn2++]=n-1;
        for(int i=0;i<k;i++)
        {
            scanf("%d %d",&x[i],&y[i]);
            sc[cn1++]=x[i];if(x[i]-1>=0)sc[cn1++]=x[i]-1;if(x[i]+1<n)sc[cn1++]=x[i]+1;
            sr[cn2++]=y[i];if(y[i]-1>=0)sr[cn2++]=y[i]-1;if(y[i]+1<n)sr[cn2++]=y[i]+1;
        }
        sort(sc,sc+cn1);sort(sr,sr+cn2);
        mn=unique(sc,sc+cn1)-sc;mm=unique(sr,sr+cn2)-sr;
        for(int i=0;i<=mn;i++)
        {
            for(int j=0;j<=mm;j++)
            {
                vis[i][j]=mp[i][j]=0;
            }
        }
        for(int i=0;i<k;i++)
        {
            mp[x_hash(x[i])][y_hash(y[i])]=1;
        }
        dfs(0,0);
        int an1=0,an2=0;
        for(int i=0;i<mn;i++)
        {
                if(i!=0)
                {
                    an1+=(sc[i-1]+1+sc[i]+1)*(sc[i]-sc[i-1]-1)*5/2;
                    an1-=sc[i]-sc[i-1]-1;
                }
                for(int j=0;j<mm;j++)
                {
                    if(sr[j]+sc[i]>=n-1)an1+=get(i,j);
                    if(j<mm-1&&sr[j]+1!=sr[j+1])
                    {
                        int tmp=n-1-sc[i];
                        if(sr[j+1]-1>=tmp)
                        {
                            if(sr[j]+1>=tmp)
                            {
                                if(sc[i]==0||sc[i]==n-1)an1+=4*(sr[j+1]-sr[j]-1);
                                else an1+=5*(sr[j+1]-sr[j]-1);
                            }
                            else
                            {
                               if(sc[i]==0||sc[i]==n-1)an1+=4*(sr[j+1]-tmp);
                                else an1+=5*(sr[j+1]-tmp);
                            }
                        }
                    }
                    an2+=get(i,j);
                }
        }
        an2+=5*(n*n-mn*mm);
        an2-=2*(n-mn)+2*(n-mm);
        int gc=__gcd(an1,an2);
        an1/=gc;an2/=gc;
        printf("Case #%d: %d/%d\n",cas++,an1,an2);
    }


    return 0;
}