UVA 11624 fire! (BFS)

fire!

Time Limit: 1000MS

Memory Limit: Unknown

64bit IO Format: %lld &%llu

Submit Status

Description

Problem b:fire! Joe works in a maze. Unfortunately, portions of the maze has 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 is on fire, you must determine whether Joe can exit the MA Ze before the fire reaches him, and what fast he can do it.

Joe and the fire all move one square per minute, vertically or horizontally (not diagonally). The fire spreads all four directions from each square, that's on fire. Joe may exit the maze from any square that borders the edge of the maze. Neither Joe nor the fire could enter a square that's occupied by a wall.

Input specificationthe first line of input contains a single integer, the number of the test cases to follow. The first line of all test case contains the integers Rand C, separated by spaces, with 1 <= R, C<= 1000. The following RLines of the test case each contain one row of the maze. Each of these lines contains exactly Ccharacters, and each of the 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's on fire

There'll be exactly oneJIn each test case. Sample Input

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

Output specificationfor Each test case, output a single line containingImpossibleIf Joe cannot exit the maze before the fire reaches him, or an integer giving the earliest time Joe can safely exit the MA Ze, in minutes. Output for Sample Input

3IMPOSSIBLE

Test Instructions:There are multiple sources of fire f,j and F can only move up and down one lattice, ask J to escape the minimum time, can not escape the output "impossible". The following:BFS, the ignition source first into the queue, and marked as having fire, then J enqueued, marked without fire, to the edge of ITCSC escape. CODE:

#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <string > #include <algorithm> #include <cstdlib> #include <set> #include <queue> #include <stack > #include <vector> #include <map> #define N 100010#define Mod 10000007#define Lson l,mid,idx<<1#  Define Rson mid+1,r,idx<<1|1#define LC idx<<1#define RC Idx<<1|1const Double EPS = 1e-11;const double PI = ACOs ( -1.0); const double E = 2.718281828;typedef long long ll;const int INF = 1000010;using namespace Std;int n,m;boo L Vis[1010][1010];char mp[1010][1010];int xx[4]= { -1,0,1,0};int yy[4]= {0,1,0,-1};int Fx,Fy,Jx,Jy;int flag;int len;    struct node{int x, y;    int num; int fire;};    Node Q[10000];queue<node>que;void BFS () {while (Que.size ()) Que.pop ();    memset (vis,0,sizeof Vis);    Node a,t;        for (int i=0; i<len; i++) {Que.push (q[i]);    Vis[q[i].x][q[i].y]=1;    } a.x=jx,a.y=jy,a.num=0,a.fire=0;Que.push (a);    Vis[jx][jy]=1;        while (Que.size ()) {A=que.front ();        Que.pop (); if (a.x==1| | a.x==n| | a.y==1| |            A.Y==M) &&a.fire==0) {cout<<a.num+1<<endl;            flag=0;        Return            } for (int i=0; i<4; i++) {t.x=xx[i]+a.x;            T.Y=YY[I]+A.Y; if (T.x>=1&&t.x<=n&&t.y>=1&&t.y<=m&&!vis[t.x][t.y]&&mp[t.x]                [t.y]!= ' # ') {t.num=a.num+1;                T.fire=a.fire;                Que.push (t);            Vis[t.x][t.y]=1;    }}}}int Main () {int T;    cin>>t;            {while (t--) {scanf ("%d%d", &n,&m);            GetChar ();            len=0; for (int i=1, i<=n; i++) {for (int j=1; j<=m; J + +) {Scan                    F ("%c", &mp[i][j]);          if (mp[i][j]== ' J ')          {jx=i,jy=j; } if (mp[i][j]== ' F ') {Q[len].x=i,q[len].y=j,q[len].num=0,q[le                    n++].fire=1;            }} getchar ();            } flag=1;            BFS ();        if (flag) cout<< "impossible\n"; }} return 0;}

UVA 11624 fire! (BFS)