bfs求单源最短路径问题.
由于bfs的层次遍历特性,最先到达终点的路径就是最短路径.
模板题值得一背＜（＾－＾）＞

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
const int N = 205;
int T,R,C;//T组用例,R行C列;
char map[N][N];//存放迷宫;
int dx[] = {-1,0,1,0},dy[] = {0,1,0,-1};//方向标;
int dist[N][N];//走到(i,j)时的步数dist[i][j];
queue<pair<int,int>> que;
int bfs (int x,int y) {
    dist[x][y] = 0;//起点设置为0;
    que.push({x,y});//将起点加入队列;
    while (!que.empty()) {
        pair<int,int> now = que.front();
        que.pop();
        for (int i = 0;i < 4;i++) {//遍历四个方向;
            int m = now.first + dx[i],n = now.second + dy[i];
            if (m < 0 || m >= R || n < 0 || n >= C) continue;//走出了迷宫的范围;
            if (dist[m][n] != -1 || map[m][n] == '#') continue;//已经走过该位置或者有障碍物;
            dist[m][n] = dist[now.first][now.second] + 1;
            if (map[m][n] == 'E')   return dist[m][n];
            que.push({m,n});//满足特判条件且不为终点的点入队列;
        }
    }
    return -1;
}
int main () {
    cin >> T;
    while(T--) {
        //初始化;
        memset(dist,-1,sizeof(dist));
        while(!que.empty()) que.pop();
        cin >> R >> C;
        int x,y;
        for (int i = 0;i < R;i++) {
            for (int j = 0;j < C;j++) {
                cin >> map[i][j];
                if (map[i][j] == 'S') {
                    x = i;
                    y = j;
                }
            }
        }
        int step = bfs(x,y);
        if (step == -1) cout << "oop!" << endl;
        else    cout << step << endl;
    }
    return 0;
}