kuangbin 简单搜索 J 双bfs

Joe works in a maze. Unfortunately, portions of the maze have

caught on fire, and the owner of the maze neglected to create a fire

escape plan. Help Joe escape the maze.

Given Joe’s location in the maze and which squares of the maze

are on fire, you must determine whether Joe can exit the maze before

the fire reaches him, and how fast he can do it.

Joe and the fire each move one square per minute, vertically or

horizontally (not diagonally). The fire spreads all four directions

from each square that is on fire. Joe may exit the maze from any

square that borders the edge of the maze. Neither Joe nor the fire

may enter a square that is occupied by a wall.

Input

The first line of input contains a single integer, the number of test

cases to follow. The first line of each test case contains the two

integers R and C, separated by spaces, with 1 ≤ R, C ≤ 1000. The

following R lines of the test case each contain one row of the maze. Each of these lines contains exactly

C characters, and each of these characters is one of:

• #, a wall

• ., a passable square

• J, Joe’s initial position in the maze, which is a passable square

• F, a square that is on fire

There will be exactly one J in each test case.

Output

For each test case, output a single line containing ‘IMPOSSIBLE’ if Joe cannot exit the maze before the

fire reaches him, or an integer giving the earliest time Joe can safely exit the maze, in minutes.

Sample Input

2
4 4
####
#JF#
#..#
#..#
3 3
###
#J.
#.F

Sample Output

3

IMPOSSIBLE

题解：

WA一次。原因是没考虑多个fire。

TLE一次。原因是一开始把所有fire起始状态存入vector，对每个fire起始状态进行bfs搜索。这样状态太多，直接把每个起始状态存入queue中，层次遍历即可。注意判断火蔓延顺序。

思路：

BFSF求出火势蔓延的顺序。

BFSJ求出J逃生的最短路，注意J去的地方必须在火势蔓延之前。

如果出不去，输出impossible.

代码：

#include <bits/stdc++.h>

using namespace std;
const int INF = 0x3f3f3f3f;
struct Node
{
int x,y;
Node(int x,int y):x(x),y(y){}
Node(){}
};

struct Status
{
int ftime,steps;
Status(int ftime,int steps):ftime(ftime),steps(steps){}
Status(){}
};

const int maxn = 1000+10;
char maze[maxn][maxn];
Status d[maxn][maxn];

int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};

int Jx,Jy;
int Fx,Fy;
int R,C;

void bfsJ()
{
queue<Node> que;
que.push(Node(Jx,Jy));
d[Jx][Jy].steps=0;

while(que.size())
{
Node s = que.front();
que.pop();

if(s.x==0||s.x==R-1||s.y==0||s.y==C-1)
{
cout<<d[s.x][s.y].steps+1<<endl;
return;
}

for(int i=0;i<4;i++)
{
int nx = s.x+dir[i][0],ny=s.y+dir[i][1];
if(nx>=0&&nx<R&&ny>=0&&ny<C&&maze[nx][ny]=='.'&&d[nx][ny].steps==INF&&d[nx][ny].ftime>d[s.x][s.y].steps+1)
{
d[nx][ny].steps=d[s.x][s.y].steps+1;
que.push(Node(nx,ny));
}
}
}
printf("IMPOSSIBLE\n");
}

queue<Node> fire;

void bfsF()
{
while(fire.size())
{
Node start = fire.front();
fire.pop();

for(int i=0;i<4;i++)
{
int nx = start.x+dir[i][0],ny=start.y+dir[i][1];

if(nx>=0&&nx<R&&ny>=0&&ny<C&&maze[nx][ny]!='#'&&d[start.x][start.y].ftime+1<d[nx][ny].ftime)
{
d[nx][ny].ftime=d[start.x][start.y].ftime+1;

fire.push(Node(nx,ny));
}
}
}
}

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
while(!fire.empty()) fire.pop();
memset(d,0x3f,sizeof(d));
cin>>R>>C;
for(int i=0;i<R;i++)
for(int j=0;j<C;j++)
{
cin>>maze[i][j];
if(maze[i][j]=='J')
{
Jx=i;
Jy=j;
}
if(maze[i][j]=='F')
{
fire.push(Node(i,j));
d[i][j].ftime=0;
}
}

bfsF();

/*for(int i=0;i<R;i++)
{
for(int j=0;j<C;j++)
{
if(d[i][j].ftime==INF)
cout<<'#'<<" ";
else
cout<<d[i][j].ftime<<" ";
}
cout<<endl;
}*/
bfsJ();
/*for(int i=0;i<R;i++)
{
for(int j=0;j<C;j++)
{

cout<<d[i][j].steps<<" ";
}
cout<<endl;
}*/

}
return 0;
}