11624 - Fire! 图论之双BFS

Joe works in a maze. Unfortunately, portions of the maze have
caught on re, and the owner of the maze neglected to create a re
escape plan. Help Joe escape the maze.
Given Joe's location in the maze and which squares of the maze
are on re, you must determine whether Joe can exit the maze before
the re reaches him, and how fast he can do it.
Joe and the re each move one square per minute, vertically or
horizontally (not diagonally). The re spreads all four directions
from each square that is on re. Joe may exit the maze from any
square that borders the edge of the maze. Neither Joe nor the re
may enter a square that is occupied by a wall.
Input
The rst line of input contains a single integer, the number of test
cases to follow. The rst 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 re
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
re 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

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//分析  看到这题目以为很好搞  好久没写 不熟练 搞了一个晚上

首先用BFS人走  book记录有没走过  题目说火不能过墙  所以要用Book记录而不能用map=# 坑  其次总体思路就是扩散  但不同于一般就是要搞一个循环将每一次进队的一个时间点所有都枚举  在代码中num循环那里  之前只记得将火每次扩散 忘了将人走的也要扩散 导致wa  意思是  

   c

  c b c

c b a b c

   c b c

    c 

  一个字母类型代表1分钟   每一个时刻要将四个方向以前进队的拿出来扩散新的进队 而不是像简单的BFS每次出一点枚举  这里是要每次4个点

第一次源点1个  第二次源点周围4个  abcd  第三次就要枚举abcd各自的四个方向  

//AC代码

#include <iostream>
#include <cstdio>
#include <queue>
#include <string.h>
using namespace std;
const int N=1000+10;
char map[N][N];
struct Node{
	int x,y,step;
};
bool book[N][N];
queue<Node> q,fire;
int next[4][2]={{0,1},{1,0},{-1,0},{0,-1}};
int n,m;
/*inline void print(){
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++)
		printf("%c ",map[i][j]);
		printf("\n");
	}
}*/
int main(){
	bool ok;
	Node node,tmp;
	int t,i,j,fx,fy;
	char c[N];
	scanf("%d",&t);
	while(t--){
		ok=0;
		memset(book,0,sizeof(book));
		scanf("%d%d",&n,&m);
		for(i=0;i<n;i++){
			scanf("%s",c);
			for(j=0;j<m;j++){
				map[i][j]=c[j];
				if(c[j]=='F'){
					node.x=i;
					node.y=j;
					fire.push(node);					
			}				
				else if(c[j]=='J'){
			//用book设置人是不是走过了 不要用#代替 不然火不能烧人 
				  book[i][j]=1;					
					node.x=i;
					node.y=j;
					node.step=0;
					q.push(node);
				}
			}			  
	}
		while(!q.empty()){
			int num=q.size();
			for(j=0;j<num;j++){
				node=q.front();
			q.pop();			
			if(map[node.x][node.y]=='F') //这个点不行了 被火烧了 
			   continue;
			for(i=0;i<4;i++){
				int a=node.x+next[i][0];
				int b=node.y+next[i][1];
				if(a<0||a>=n||b<0||b>=m)
				{
					ok=1;
					break;
				}
				//墙和火不能走  忘了+火 超时了 
	     	//用book设置人是不是走过了 不要用#代替 不然火不能烧人 
			 if(map[a][b]=='#'||map[a][b]=='F'||book[a][b]==1) 
			      continue;
			    book[a][b]=1;    
			     tmp.x=a;
			     tmp.y=b;
			     tmp.step=node.step+1;
			     q.push(tmp);
			}
			}
			if(ok)
			break;
			num=fire.size();
			for(j=0;j<num;j++){
				node=fire.front();
				fire.pop();
					for(i=0;i<4;i++){
						int a=node.x+next[i][0];
			         	int b=node.y+next[i][1];
			         	//题目要求障碍物 墙 不能被烧 郁闷 害我wa 
			         	if(a<0||a>=n||b<0||b>=m||map[a][b]=='#'||map[a][b]=='F') 
			         	continue;
			         	map[a][b]='F';
			            tmp.x=a;
			            tmp.y=b;
			            fire.push(tmp);
					}
			}
		//	print();
		}
		if(ok)
		  printf("%d\n",node.step+1);
		else
		  printf("IMPOSSIBLE\n");
		  while(!q.empty()) q.pop();
		  while(!fire.empty()) fire.pop();
	}
	return 0;
}