Fire

传送门UVA11624

描述

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.

输入

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:

1
2
3
4
1. #, a wall
2. ., a passable square
3. J, Joe’s initial position in the maze, which is a passable square
4. F, a square that is on fire

There will be exactly one J in each test case.

输出

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.

样例

  • Input

    1
    2
    3
    4
    5
    6
    7
    8
    9
    10
    2
    4 4
    ####
    #JF#
    #..#
    #..#
    3 3
    ###
    #J.
    #.F
  • Output

    1
    2
    3
    IMPOSSIBLE

题解

  • 题意:迷宫逃脱,一开始有多个火焰源,火焰每单位时间向四周蔓延一格,人每单位可以向四周移动一格,人可以从任意边界逃脱,但是不能碰到火焰,求最短逃脱时间。
  • 先多起点bfs预处理出每一个格子的火焰到达时间,然后对人进行bfs即可。

Code

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#include<stdio.h>
#include<string.h>
#include<queue>
#define INIT(a,b) memset(a,b,sizeof(a))
#define LL long long
using namespace std;
const int inf=0x3f3f3f3f;
const int N=1e3+7;
const int mod=1e9+7;
int t,n,m,jx,jy;
char mp[N][N];
int fi[N][N],vis[N][N];
int rec[4][2]={1,0,-1,0,0,1,0,-1};
struct node{
int x,y,time;
};
inline bool away(int i,int j){
if(i==0||i==n-1||j==0||j==m-1)return true;
else return false;
}
inline bool judge(int i,int j){
if(i<0||i>=n||j<0||j>=m||vis[i][j]||mp[i][j]=='#')return false;
return true;
}
queue<node> fque;
void fire(){
while(!fque.empty()){
node q=fque.front();fque.pop();
for(int i=0;i<4;i++){
int nx=q.x+rec[i][0],ny=q.y+rec[i][1];
if(judge(nx,ny)){
fque.push((node){nx,ny,q.time+1});
fi[nx][ny]=q.time+1;
vis[nx][ny]=1;
}
}
}
}
int bfs(){
INIT(vis,0);
queue<node> que;
que.push((node){jx,jy,1});
vis[jx][jy]=1;
while(!que.empty()){
node q=que.front();que.pop();
if(away(q.x,q.y)) return q.time;
for(int i=0;i<4;i++){
int nx=q.x+rec[i][0],ny=q.y+rec[i][1];
if(judge(nx,ny)&&q.time+1<fi[nx][ny]){
que.push((node){nx,ny,q.time+1});
vis[nx][ny]=1;
}
}
}
return -1;
}
int main(){
scanf("%d",&t);
while(t--){
INIT(fi,inf),INIT(vis,0);
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++){
getchar();
for(int j=0;j<m;j++){
scanf("%c",&mp[i][j]);
if(mp[i][j]=='J') jx=i,jy=j;
if(mp[i][j]=='F') {
fque.push((node){i,j,1});
vis[i][j]=1;
fi[i][j]=1;
}
}
}
fire();
int ans=bfs();
if(ans==-1) printf("IMPOSSIBLE\n");
else printf("%d\n",ans);
}

return 0;
}
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