Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem B: Fire!

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 Specification

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.

Sample Input

2
4 4
####
#JF#
#..#
#..#
3 3
###
#J.
#.F

Output Specification

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.

Output for Sample Input

3
IMPOSSIBLE

---

我忘記也有火到達不了的地方，但是為了節省空間，我忘了我的 F[i][j] 是 0，導致判斷錯誤。

#include <stdio.h>
#include <queue>
using namespace std;
char g[1010][1010];
int F[1010][1010], J[1010][1010];
int n, m, dir[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};
struct pt {
    int x, y;
};
pt pf, pj;
queue<pt> Q;
void sol() {
    int i, j;
    while(!Q.empty())   Q.pop();
    pt tn, tx;
    for(i = 0; i < n; i++)
        for(j = 0; j < m; j++)
            if(g[i][j] == 'F') {
                pf.x = i, pf.y = j;
                Q.push(pf);
                F[pf.x][pf.y] = 1;
            }
    while(!Q.empty()) {
        tn = Q.front();
        Q.pop();
        for(i = 0; i < 4; i++) {
            tx.x = tn.x+dir[i][0];
            tx.y = tn.y+dir[i][1];
            if(tx.x < 0 || tx.x >= n || tx.y < 0 || tx.y >= m)
                continue;
            if(g[tx.x][tx.y] == '#')
                continue;
            if(F[tx.x][tx.y] == 0) {
                F[tx.x][tx.y] = F[tn.x][tn.y]+1;
                Q.push(tx);
            }
        }
    }
    Q.push(pj), J[pj.x][pj.y] = 1;
    while(!Q.empty()) {
        tn = Q.front();
        Q.pop();
        if(tn.x == 0 || tn.y == 0 || tn.x == n-1 || tn.y == m-1) {
            printf("%d\n", J[tn.x][tn.y]);
            return;
        }
        for(i = 0; i < 4; i++) {
            tx.x = tn.x+dir[i][0];
            tx.y = tn.y+dir[i][1];
            if(tx.x < 0 || tx.x >= n || tx.y < 0 || tx.y >= m)
                continue;
            if(g[tx.x][tx.y] == '#')
                continue;
            if(J[tx.x][tx.y] == 0 && (J[tn.x][tn.y]+1 < F[tx.x][tx.y] || F[tx.x][tx.y] == 0)) {
                J[tx.x][tx.y] = J[tn.x][tn.y]+1;
                Q.push(tx);
            }
        }
    }
    puts("IMPOSSIBLE");
}
int main() {
    int t, i, j;
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &n, &m);
        for(i = 0; i < n; i++) {
            scanf("%s", g[i]);
            for(j = 0; j < m; j++) {
                if(g[i][j] == 'J')
                    pj.x = i, pj.y = j;
                F[i][j] = 0, J[i][j] = 0;
            }
        }
        sol();
    }
    return 0;
}