UVA - 11624：Fire!

Fire!

来源：UVA

标签：图论

参考资料：

相似题目：

题目

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:

• #, 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.

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.

输入样例

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

输出样例

3
IMPOSSIBLE

解题思路

用BFS先处理火到达每个地方的时间。

参考代码

#include
#include 
#include
#define MAXN 1000
using namespace std;
char maze[MAXN+5][MAXN+5];
int firetime[MAXN+5][MAXN+5];//init -1 each time
int arrivetime[MAXN+5][MAXN+5];
int vis[MAXN+5][MAXN+5];//init 0 each time
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
int row,col;
queue<int> qx,qy;void hasfire(){while(!qx.empty()){int x=qx.front();int y=qy.front();qx.pop();qy.pop();for(int i=0;i<4;i++){int nx=x+dx[i];int ny=y+dy[i];if(nx>=0 && nx<row && ny>=0 && ny<col && !vis[nx][ny] && maze[nx][ny]!='#' ){vis[nx][ny]=1;firetime[nx][ny]=firetime[x][y]+1;qx.push(nx);qy.push(ny);}}}
}int escape(){while(!qx.empty()){int x=qx.front();int y=qy.front();qx.pop();qy.pop();if(x==0 || x==row-1 || y==0 || y==col-1){return arrivetime[x][y]+1;}for(int i=0;i<4;i++){int nx=x+dx[i];int ny=y+dy[i];if(nx>=0 && nx<row && ny>=0 && ny<col && !vis[nx][ny] && maze[nx][ny]=='.' ){vis[nx][ny]=1;if(firetime[nx][ny]==-1 || arrivetime[x][y]+1<firetime[nx][ny]){arrivetime[nx][ny]=arrivetime[x][y]+1;qx.push(nx);qy.push(ny);}}}}return -1;
}void clearqueue(){while(!qx.empty()){qx.pop();qy.pop();}
}int main(){int T;scanf("%d",&T);while(T--){memset(firetime,-1,sizeof(firetime));memset(arrivetime,-1,sizeof(arrivetime));memset(vis,0,sizeof(vis));clearqueue();scanf("%d%d",&row,&col);for(int i=0;i<row;i++){scanf("%s",maze[i]);}int jx,jy;for(int i=0;i<row;i++){for(int j=0;j<col;j++){if(maze[i][j]=='F'){qx.push(i);qy.push(j);firetime[i][j]=0;vis[i][j]=1;}if(maze[i][j]=='J'){jx=i;jy=j;}}}hasfire();memset(vis,0,sizeof(vis));clearqueue();qx.push(jx);qy.push(jy);arrivetime[jx][jy]=0;vis[jx][jy]=1;int ans=escape();if(ans==-1){printf("IMPOSSIBLE\n");}else{printf("%d\n",ans);}}return 0;
}

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