题目链接I - A计划
思路：
还是正常的bfs,多了一层而已。在移动时，增加判断，如果下一步为时光传输机且对应的节点为空地，则加入队列，否则continue即可。
代码：
#include <stdio.h>
#include <iostream>
#include <vector>
#include <math.h>
#include <algorithm>
#include <queue>
#include <string.h>
#include <set>
#include <stack>
#include <stdlib.h>
#include <time.h>
using namespace std;
int mp[2][12][12];
typedef struct Node
{
int x, y, z;
int num;
}node;
int dx[] = {0,0,1,-1};
int dy[] = {1,-1,0,0};
bool bfs(int x, int y, int z, int time)
{
node temp;
temp.x = 1;
temp.y = 1;
temp.z = 0;
temp.num = 0;
queue<node> q;
q.push(temp);
while(!q.empty())
{
temp = q.front();
//cout<<temp.x<<" "<<temp.y<<" "<<temp.z<<" "<<temp.num<<endl;
if(temp.num > time)
break;
if(temp.x == x && temp.y == y && temp.z == z)
return true;
q.pop();
for(int i=0;i<4;i++)
{
node t;
t.x = temp.x + dx[i];
t.y = temp.y + dy[i];
t.z = temp.z;
t.num = temp.num + 1;
if(mp[t.z][t.x][t.y] == 2)
{
mp[t.z][t.x][t.y] = 0;
if(mp[!t.z][t.x][t.y] == 1)
{
mp[!t.z][t.x][t.y] = 0;
t.z ^= 1;
q.push(t);
}
else
continue;
}
else if(mp[t.z][t.x][t.y] == 1)
{
mp[t.z][t.x][t.y]--;
q.push(t);
}
}
}
return false;
}
int main()
{
int T;
cin>>T;
while(T--)
{
int n, m, time, sx, sy, sz;
cin>>n>>m>>time;
char t[12];
memset(mp, 0, sizeof(mp));
for(int ti=0;ti<=1;ti++)
for(int i=1;i<=n;i++)
{
cin>>t;
for(int j=0;j<m;j++)
if(t[j] == 'P')
{
sx = i;
sy = j+1;
sz = ti;
mp[ti][i][j+1] = 1;
}
else if(t[j] == '#')
mp[ti][i][j+1] = 2;
else if(t[j] == '.')
mp[ti][i][j+1] = 1;
}
if(bfs(sx, sy, sz, time))
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}