【题解】1215：迷宫

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#define N 101
using namespace std;
char ch;
int n,xa,ya,xb,yb;
bool flag;
int vis[N][N];
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
void dfs(int x,int y)
{
    for(int i=0;i<4;i++)
    {
        int nx=x+dir[i][0];
        int ny=y+dir[i][1];
        if(nx>=0&&nx<n&&ny>=0&&ny<n&&vis[nx][ny]==0)
        {
            vis[nx][ny]=1;
            if(nx==xb&&ny==yb)
            {
                cout<<"YES"<<endl;
                flag=true;
                break;
            }
            else
                dfs(nx,ny);
        }
    }
}
int main()
{
    int k;
    cin>>k;
    while(k--)
    {
        memset(vis,0,sizeof(vis));
        flag=false;
        cin>>n;
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
            {
                cin>>ch;
                if(ch=='#')
                    vis[i][j]=1;
            }
        cin>>xa>>ya;
        cin>>xb>>yb;
        if(vis[xa][ya]||vis[xb][yb])
        {
            cout<<"NO"<<endl;
            continue;
        }
        else
            dfs(xa,ya);
        if(!flag)
            cout<<"NO"<<endl;
    }
    return 0;
}

解法2

#include <bits/stdc++.h>
using namespace std;
#define N 105
int n, ha, la, hb, lb;
char mp[N][N];
bool vis[N][N];
int dir[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
bool isReach;
void dfs(int sx, int sy)
{
    if(isReach)//剪枝：如果发现已经能到了，就不再递归运行了 
        return;
    for(int i = 0; i < 4; ++i)
    {
        int x = sx + dir[i][0], y = sy + dir[i][1];
        if(x >= 0 && x < n && y >= 0 && y < n && vis[x][y] == false && mp[x][y] == '.')
        {
            vis[x][y] = true;
            if(x == hb && y == lb)
            {
                isReach = true;
                return;    
            }
            else
                dfs(x, y);
        }
    }
}
int main()
{//判断起点终点是否在一个连通块内 
    int k;
    cin >> k;
    while(k--)
    {
        cin >> n;
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j)
                cin >> mp[i][j];
        cin >> ha >> la >> hb >> lb;
        memset(vis, 0, sizeof(vis));
        isReach = false;
        vis[ha][la] = true;
        if(ha == hb && la == lb)//如果起点和终点重合，直接就能到达 
            isReach = true;
        if(mp[ha][la] != '#' && mp[hb][lb] != '#')
            dfs(ha, la);
        cout << (isReach ? "YES" : "NO") << endl;
    }
	return 0;
}