Fire！ UVA-11624

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

这道题没什么好说的，一个最短路径的题，用宽搜。但是要注意的是，这道题是两个宽搜，先算火（有多个火！）到每个点的距离，然后人就必须在火到那个点之前到达那个点，不然人（只有一个人）就要死了

#include<iostream>#include<cmath>#include<queue>#include<cstring>#include<queue>using namespace  std; const int maxn=1006;const int inf=0x3f3f3f3f;char a[maxn][maxn];int man[maxn][maxn];int fire[maxn][maxn];int direction[4][2]={{0,1},{1,0},{-1,0},{0,-1}};int line,column;int start,start1;struct wo{int x;int y;};queue<wo>p;int makefire(){//着火//while(p.size())p.pop()在main函数里面已经清空了;while(p.size()){wo help1;help1=p.front();p.pop();for(int i=0;i<4;i++){int x1=help1.x+direction[i][0],y1=help1.y+direction[i][1]; if(x1<1||x1>line)continue;if(y1<1||y1>column)continue;if(a[x1][y1]=='#')continue;if(fire[x1][y1]!=inf)continue;//没越出边界而且还不是墙且没走过才行fire[x1][y1]=fire[help1.x][help1.y]+1;wo help2;help2.x=x1;help2.y=y1;
p.push(help2);}}}int run(){while(p.size())p.pop();//清空    wo ps;ps.x=start;ps.y=start1;    p.push(ps);    while(p.size()){         ps=p.front();p.pop();         if(ps.x==1||ps.x==line||ps.y==1||ps.y==column){         //cout<<ps.x<<" "<<ps.y<<endl;         return man[ps.x][ps.y]+1; }for(int i=0;i<4;i++){int x1=ps.x+direction[i][0],y1=ps.y+direction[i][1];if(x1<1||x1>line)continue;if(y1<1||y1>column)continue;if(man[x1][y1]<=man[ps.x][ps.y]+1)continue;if(a[x1][y1]=='#')continue;if(fire[x1][y1]<=man[ps.x][ps.y]+1)continue;//当时那个点没着火且不是墙，没走过才行（重走你走过的点无聊吗？）man[x1][y1]=man[ps.x][ps.y]+1;wo help2;help2.x=x1;help2.y=y1;p.push(help2);}
} return -1;}int main(){int t;cin>>t;while(t--){memset(man,inf,sizeof(man));//初始化memset(fire,inf,sizeof(fire));cin>>line>>column;wo help;for(int i=1;i<=line;i++){for(int j=1;j<=column;j++){            cin>>a[i][j];if(a[i][j]=='J'){start=i;start1=j;man[start][start1]=0;}else if(a[i][j]=='F'){help.x=i;help.y=j;fire[i][j]=0;p.push(help);//入队}}}makefire();//圣火int ff=-1;
ff=run();if(ff==-1)cout<<"IMPOSSIBLE"<<endl;跑不出去就只有乖乖等死了else cout<<ff<<endl;while(p.size())p.pop();}}