Java(javac 1.8) 解法, 执行用时: 395ms, 内存消耗: 31384K, 提交时间: 2020-05-30 11:17:33

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		try (Scanner sc = new Scanner(System.in)) {
			int N = sc.nextInt(), a = 0, b = 0;
			boolean[][] grid = new boolean[100][100];
			for (int i = 0; i < N; i++) {
				grid[sc.nextInt() - 1][sc.nextInt() - 1] = true;
			}
			for (int i = 0; i < 100; i++) {
				for (int j = 0; j < 100; j++) {
					if (dfs(i, j, grid)) {
						dfs(grid, i, j);
					}
				}
			}
			for (int i = 0; i < 100; i++) {
				for (int j = 0; j < 100; j++) {
					if (grid[i][j]) {
						a++;
						b += (i < 99 && grid[i + 1][j] ? 1 : 0) + (j < 99 && grid[i][j + 1] ? 1 : 0);
					}
				}
			}
			System.out.println(4 * a - 2 * b);
		}
	}

	private static boolean dfs(int i, int j, boolean[][] grid) {
		if (i < 0 || i > 99 || j < 0 || j > 99) {
			return false;
		} else if (grid[i][j]) {
			return true;
		}
		grid[i][j] = true;
		boolean result = dfs(i - 1, j, grid) && dfs(i, j - 1, grid) && dfs(i, j + 1, grid) && dfs(i + 1, j, grid);
		grid[i][j] = false;
		return result;
	}

	private static void dfs(boolean[][] grid, int i, int j) {
		if (i < 0 || i > 99 || j < 0 || j > 99) {
			return;
		} else if (!grid[i][j]) {
			grid[i][j] = true;
			dfs(grid, i - 1, j);
			dfs(grid, i, j - 1);
			dfs(grid, i, j + 1);
			dfs(grid, i + 1, j);
		}
	}
}

C++14(g++5.4) 解法, 执行用时: 9ms, 内存消耗: 992K, 提交时间: 2020-06-22 23:52:56

#include<bits/stdc++.h>
using namespace std;
set< pair<int,int> > st;
set< pair<int, int> > vis;
int n;
int ans = 0;
int dx[]={0,0,-1,1,1,1,-1,-1};
int dy[]={1,-1,0,0,1,-1,1,-1};

bool check(int x, int y)
{
    for(int i = 0; i < 8; i++) if(st.count(make_pair(x+dx[i], y+dy[i]))) return 1;
    return 0;
}

void dfs(int x, int y)
{
    if(st.count(make_pair(x, y)) )
    {
        ans++;
        return;
    }
    if(vis.count(make_pair(x, y)) )
    {
        return;
    }
    if(!check(x, y)) return;
    vis.emplace(make_pair(x, y));
    for(int i = 0; i < 4; i++) dfs(x+dx[i], y+dy[i]);
}
int main()
{
    scanf("%d",&n);
    for(int i = 0, x, y ; i < n; i++) scanf("%d %d",&x,&y), st.emplace(make_pair(x, y));
    auto p = *st.begin();
    dfs(p.first-1, p.second);
    printf("%d\n",ans);
}

C++11(clang++ 3.9) 解法, 执行用时: 16ms, 内存消耗: 864K, 提交时间: 2020-05-30 10:47:37

#include<bits/stdc++.h>
#define Max 1000000
using namespace std;
int n,y,x,i,sy=Max,sx=Max,ans;
const int dx[4]={0,1,0,-1};
const int dy[4]={1,0,-1,0};
map<int,map<int,int> >mat;
int main(){
	cin>>n;
	for(int j=1;j<=n;j++){
		cin>>x>>y;
		mat[y][x]=1;
		if(y<sy||(y==sy&&x<sx)) sy=y,sx=x;
	}
	x=sx;y=sy;
	do{
		x+=dx[i];y+=dy[i];ans++;
		for(i=(i+3)%4; ;i=(i+1)%4)
			if((i==0&&mat[y][x])||(i==1&&mat[y-1][x])||(i==2&&mat[y-1][x-1])||(i==3&&mat[y][x-1])) break;//如果我们已经搜索到的话，跳出即可。
	}while(x!=sx||y!=sy);
	cout<<ans<<endl;
	return 0;
}