题意：
给定n*m 大小矩阵， 出发点所在列， 终点所在列 (行统一在第一行和最后一行)
下面 2*i - 1 行给出 矩阵中墙壁存在与否
问沿着右边墙壁走，从起点走到终点 ，再从终点走到起点能否遍历所有方格
思路：
BFS模拟题意，队列存边
#include<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
#include<algorithm>
#include<set>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <math.h>
#include <queue>
#define N 1500
#define ll double
using namespace std;
inline ll Min(ll a,ll b){return a>b?b:a;}
inline ll Max(ll a,ll b){return a>b?a:b;}
int map[N][N];
int n, m, enter, out;
struct node{
int x,y,from;
node(int a=0,int b=0,int c=0):x(a),y(b),from(c){}
};
int go(int x){// turn right
if(x==1)return 4;
if(x==2)return 1;
if(x==3)return 2;
if(x==4)return 3;
}
int go2(int x){// turn left
if(x==1)return 2;
if(x==2)return 3;
if(x==3)return 4;
if(x==4)return 1;
}
bool Inmap(int x,int y){
if(x<0 || y<0 || x>=(n<<1) || y>=(m<<1))return false;
return true;
}
bool hehe(node a){//could turn right
if(a.from==1)
{
a.from = 4;
if( Inmap(a.x,a.y+2) && !map[a.x][a.y+1])
return true;
return false;
}
if(a.from==2)
{
a.from = 1;
if( Inmap(a.x-2,a.y) && !map[a.x-1][a.y])
return true;
return false;
}
if(a.from==3)
{
a.from = 2;
if( Inmap(a.x,a.y-2) && !map[a.x][a.y-1])
return true;
return false;
}
if(a.from==4)
{
a.from = 3;
if( Inmap(a.x+2,a.y) && !map[a.x+1][a.y])
return true;
return false;
}
}
node zz(node a){//after turn right,the coordinate
if(a.from==1)
{
a.from = 4;
if( Inmap(a.x,a.y+2) && !map[a.x][a.y+1])
a.y+=2;;
return a;
}
if(a.from==2)
{
a.from = 1;
if( Inmap(a.x-2,a.y) && !map[a.x-1][a.y])
a.x-=2;
return a;
}
if(a.from==3)
{
a.from = 2;
if( Inmap(a.x,a.y-2) && !map[a.x][a.y-1])
a.y-=2;
return a;
}
if(a.from==4)
{
a.from = 3;
if( Inmap(a.x+2,a.y) && !map[a.x+1][a.y])
a.x+=2;
return a;
}
}
int vis[N][N],papa[N][N]; //papa是记录是否访问过该点 vis是BFS后这个访问的点
void BFS(node hehehe, node endd){ // hehehe为起点 endd为终点
memset(vis,0,sizeof(vis));
queue<node>q;
q.push(hehehe);
while(!q.empty()){
node E = q.front(); q.pop();
vis[E.x][E.y] ++;
if(E.x == endd.x && E.y == endd.y && E.from == endd.from)return; //注意走到终点后方向也必须是确定的
if(hehe(E))
{
node now = zz(E);
if(vis[now.x][now.y]<=2){q.push(now);continue;}
}
E.from = go2(E.from);
if(hehe(E))
{
node now = zz(E);
if(vis[now.x][now.y]<=2){q.push(now);continue;}
}
E.from = go2(E.from);
if(hehe(E))
{
node now = zz(E);
if(vis[now.x][now.y]<=2){q.push(now);continue;}
}
E.from = go2(E.from);
if(hehe(E))
{
node now = zz(E);
if(vis[now.x][now.y]<=2){q.push(now);continue;}
}
}
}
int main(){
int T, i, j, temp;scanf("%d",&T);
while(T--){
memset(map,0,sizeof(map));
scanf("%d %d %d %d",&n,&m,&enter,&out);
enter++, out++;
for(i = 1; i< (2*n); i++)
{
if(i&1)
{
temp = m-1;
for(j = 1; temp-- ;j+=2)
scanf("%d",&map[i][j]);
}
else
{
temp = m;
for(j = 0; temp--; j+=2)
{
scanf("%d",&map[i][j]);
if(temp)
map[i][j+1]=1;
}
}
}
memset(papa,0,sizeof(papa));
BFS(node(1,(enter-1)<<1,3), node(1+((n-1)<<1),(out-1)<<1,4));
for(i = 1; i < (n<<1); i+=2)for(j = 0; j < (m<<1); j+=2)if(vis[i][j]>0) papa[i][j]=1;
BFS( node(1+((n-1)<<1),(out-1)<<1,1), node(1,(enter-1)<<1,2));
for(i = 1; i < (n<<1); i+=2) for(j = 0; j < (m<<1); j+=2)if(vis[i][j]>0)
papa[i][j]=1;
bool ok = true;
for(i = 1; i < (n<<1) && ok; i+=2)
for(j = 0; j < (m<<1) && ok; j+=2)
if(!papa[i][j]) ok = false;
if(ok)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
/*
99
3 4 1 2
0 0 0
1 1 0 1
0 0 0
0 0 0 0
1 0 0
2 2 1 2
0
1 0
0
2 2 1 2
0
1 0
0
2 2 1 2
0
0 1
0
3 3 2 2
0 0
0 1 0
1 1
0 1 0
0 0
3 3 2 2
0 0
0 0 0
0 0
0 0 0
0 0
4 3 1 3
0 0
0 0 0
0 0
0 0 0
0 0
0 0 0
0 0
4 3 1 2
0 0
0 1 0
1 1
0 0 0
1 1
0 0 0
1 0
4 3 1 2
0 0
0 1 0
1 0
0 0 1
1 0
0 0 0
1 0
ans:
y
y
y
y
n
n
n
y
y
*/
