| Time Limit: 5000MS | Memory Limit: 262144KB | 64bit IO Format: %lld & %llu |
Description
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 the two integers R and C, separated by spaces, with 1 <= R, C <= 1000. The following R lines of input 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
Output
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
Case #1: 4 4 #### #JF# #..# #..# Case #2: 3 3 ### #J. #.F
Sample Output
3 IMPOSSIBLE
Source
waterloo 13 June, 2009
題目本身不難，但是很容易WA，兩個坑點：一開始就在邊界的情況和火堆根本碰不到人的情況，第一種稍微把BFS判斷順序改一下就可以了，第二種會導致比較人和火誰先跑出迷宮這種BFS方法WA，可以假設若火根本碰不到人，那麼就算火跑的比香港記者更快，那人還是可以走出去的，然後怎麼保證每一個地方都是最早被火碰到的呢？把所有的火都壓入隊列一次性BFS掉，這樣保證層數相同時時間可以統一更新，因此要用另一種保險的BFS方法，把每一個點時間設好，再對人進行BFS，被這兩個坑點弄的WA十餘次………哎還是too naive。哦對了題目本身是PDF但是CSDN很難放PDF轉換爲word也很難看，乾脆找了個只是輸出格式修改了的題面放上去，方便起見自己把輸出格式也改成了原題一樣的……
代碼：
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define MM(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=1010;
struct info
{
int x;
int y;
int t;
info operator+(info b)
{
b.x+=x;
b.y+=y;
b.t+=t;
return b;
}
};
info J,F,direct[4]={{1,0,1},{-1,0,1},{0,1,1},{0,-1,1}};
char pos[N][N];
int firetime[N][N];
int vis[N][N];
int n,m;
queue<info>Q;
inline bool checkfire(const info &f)
{
return (f.x>=0 && f.x<n && f.y>=0 && f.y<m && pos[f.x][f.y]!='#' && vis[f.x][f.y]==0);
}
inline bool checkman(const info &a)
{
return (a.x>=0 && a.x<n && a.y>=0 && a.y<m && pos[a.x][a.y]!='#' && pos[a.x][a.y]!='F' && vis[a.x][a.y]==0);
}
inline void init()
{
MM(pos,0);
MM(firetime,INF);
J.x=-1,J.y=-1,J.t=-1;
while (!Q.empty())
Q.pop();
}
inline void setfire()
{
MM(vis,0);
while (!Q.empty())
{
info now=Q.front();
Q.pop();
for (int i=0; i<4; ++i)
{
info v=now+direct[i];
if(checkfire(v))
{
vis[v.x][v.y]=1;
firetime[v.x][v.y]=v.t;
Q.push(v);
}
}
}
}
inline int run(const info &s)
{
MM(vis,0);
queue<info>q;
vis[s.x][s.y]=1;
q.push(s);
while (!q.empty())
{
info now=q.front();
q.pop();
if(now.x==0||now.x==n-1||now.y==0||now.y==m-1)
return now.t;
for (int i=0; i<4; ++i)
{
info v=now+direct[i];
if(checkman(v)&&v.t<firetime[v.x][v.y])
{
vis[v.x][v.y]=1;
q.push(v);
}
}
}
return -1;
}
int main(void)
{
int tcase,i,j,out,cnt;
scanf("%d",&tcase);
while (tcase--)
{
init();
cnt=0;
out=-1;
scanf("%d%d",&n,&m);
for (i=0; i<n; ++i)
{
scanf("%s",pos[i]);
for (j=0; j<m; ++j)
{
if(pos[i][j]=='J')
{
J.x=i;
J.y=j;
J.t=0;
}
else if(pos[i][j]=='F')
{
F.x=i;
F.y=j;
F.t=0;
Q.push(F);
}
}
}
setfire();
out=run(J);
out==-1?puts("IMPOSSIBLE"):printf("%d\n",out+1);
}
return 0;
}