Fire! （雙bfs+預處理）
題目鏈接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2671
題目：
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.
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.
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 IMPOSSIBL
題意：Joe要逃離著火的森林，Joe和火都只能往上下左右四個方向轉移，Joe到達邊界即可離開，問最小逃離步數，如果不能輸出IMPOSSIBL。坑點：是逃離後的步數，因而要在到達邊界的步數上+1，火有多個（因為這個WA了好久==!）。
思路：先用t數組預處理出火到達每個地方的時間，然後再對Joe進行bfs即可。
代碼實現如下：
1 #include <queue> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 6 const int inf = 0x3f3f3f3f; 7 int T, r, c, ans, sx, sy; 8 char mp[1007][1007]; 9 int vis[1007][1007], t[1007][1007]; 10 11 struct node{ 12 int x, y; 13 int step; 14 }nw, nxt; 15 16 int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1}; 17 18 19 void bfs1() { 20 queue<node> q; 21 for(int i = 0 ; i< r; i++) { 22 for(int j = 0; j < c; j++) { 23 if(mp[i][j] == ‘F‘) { 24 nw.x = i, nw.y = j; 25 t[i][j] = 0; 26 q.push(nw); 27 } 28 } 29 } 30 while(!q.empty()) { 31 nw = q.front(), q.pop(); 32 for(int i = 0; i < 4; i++) { 33 nxt.x = nw.x + dx[i], nxt.y = nw.y + dy[i]; 34 if(nxt.x >= 0 && nxt.x < r && nxt.y >= 0 && nxt.y < c && mp[nxt.x][nxt.y] != ‘#‘ && t[nxt.x][nxt.y] > t[nw.x][nw.y] + 1) { 35 t[nxt.x][nxt.y] = t[nw.x][nw.y] + 1; 36 q.push(nxt); 37 } 38 } 39 } 40 } 41 42 void bfs2(int x, int y) { 43 nw.x = x, nw.y = y, nw.step = 0; 44 vis[x][y] = 1; 45 queue<node> q; 46 q.push(nw); 47 while(!q.empty()) { 48 nw = q.front(), q.pop(); 49 if(nw.x ==0 || nw.x == r - 1 || nw.y == 0 || nw.y == c - 1) { 50 ans = nw.step + 1; 51 return; 52 } 53 for(int i = 0; i < 4; i++) { 54 nxt.x = nw.x + dx[i], nxt.y = nw.y + dy[i]; 55 if(nxt.x >= 0 && nxt.x < r && nxt.y >= 0 && nxt.y <c && mp[nxt.x][nxt.y] != ‘#‘ && nw.step + 1 < t[nxt.x][nxt.y] && vis[nxt.x][nxt.y] == 0) { 56 vis[nxt.x][nxt.y] = 1; 57 nxt.step = nw.step + 1; 58 q.push(nxt); 59 } 60 } 61 } 62 } 63 64 int main() { 65 scanf("%d", &T); 66 while(T--) { 67 scanf("%d%d", &r, &c); 68 for(int i = 0; i < r; i++) { 69 scanf("%s", mp[i]); 70 for(int j = 0; j < c; j++) { 71 if(mp[i][j] == ‘J‘) { 72 sx = i, sy = j; 73 } 74 } 75 } 76 memset(vis, 0, sizeof(vis)); 77 memset(t, inf, sizeof(t)); 78 ans = inf; 79 bfs1(); 80 bfs2(sx, sy); 81 if(ans >= inf) printf("IMPOSSIBLE\n"); 82 else printf("%d\n", ans); 83 } 84 return 0; 85 }
Fire! （雙bfs+預處理）