Fire!
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
题目简述:
J困在迷宫里,迷宫里可能有火,火每秒向上下左右蔓延一格,但不能烧到墙上,问小J能不能成功跑出来。
题目分析:
处理一下每个格子着火的时间,初始化为无穷,着火后不能踩上去。BFS即可。
代码实现:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<stack>
#include<cstdlib>
#include<cmath>
#include<queue>
//#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
#define pf printf
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define ms(i,j) memset(i,j,sizeof(i))
int n,m,fire[1005][1005],d[4][2]= {-1,0,1,0,0,-1,0,1};
char a[1005][1005];
bool mark[1005][1005];
struct sb
{
int x,y,s;
} tmp,now,bgn;
bool valid()
{
return tmp.x>=0&&tmp.x<n&&tmp.y>=0&&tmp.y<m;
}
bool ok()
{
if(now.x==0||now.y==0||now.x==n-1||now.y==m-1)
return 1;
else
return 0;
}
void bfs()
{
queue<sb> line;
ms(mark,0);
line.push(bgn);
mark[bgn.x][bgn.y]=1;
while(!line.empty())
{
now=line.front();
line.pop();
if(ok())
{
pf("%d\n",now.s+1);
return;
}
for(int i=0; i<4; i++)
{
tmp.x=now.x+d[i][0];
tmp.y=now.y+d[i][1];
tmp.s=now.s+1;
if(valid()&&!mark[tmp.x][tmp.y]&&tmp.s<fire[tmp.x][tmp.y]&&a[tmp.x][tmp.y]!='#')
{
mark[tmp.x][tmp.y]=1;
line.push(tmp);
}
}
}
pf("IMPOSSIBLE\n");
}
int main()
{
int T;
queue<sb> q;
scanf("%d",&T);
while(T--)
{
ms(mark,0);
fill(fire[0],fire[0]+1005*1005,INF);
ms(a,0);
sff(n,m);
for(int i=0; i<n; i++)
{
scanf("%s",a[i]);
for(int j=0; j<m; j++)
{
if(a[i][j]=='J')
bgn= {i,j,0};
if(a[i][j]=='F')
{
q.push(tmp= {i,j,0});
mark[i][j]=1;
fire[i][j]=0;
}
}
}
while(!q.empty())
{
now=q.front();
q.pop();
for(int i=0; i<4; i++)
{
tmp.x=now.x+d[i][0];
tmp.y=now.y+d[i][1];
tmp.s=now.s+1;
if(valid()&&!mark[tmp.x][tmp.y]&&a[tmp.x][tmp.y]!='#')
{
mark[tmp.x][tmp.y]=1;
fire[tmp.x][tmp.y]=tmp.s;
q.push(tmp);
}
}
}
bfs();
}
}