hdoj 1242 Rescue

http://acm.hdu.edu.cn/showproblem.php?pid=1242

//搜索题：广度优先搜索/*当前格视为节点，其上下左右视为下一层，用广度优先搜索，将每层的节点入队。对已经访问的格子标记已经访问。节点附带属性为坐标和时间，由于有守卫影响，所以同一个格子可能出现不同的时间属性，故将守卫的格子视为两层，对守卫的格子进行标记，以确定是否为第一次访问将其入队两次。最后，访问到目的地时候break退出*/#include <iostream>#include <queue>using namespace std;struct  Pos{int i, j, time;};int main(){char M[200][200]; // 监狱int n, m;int time, bi, bj, i, j;bool visited[200][200], x[200][200], flag;int dir[4][2] = {{-1,0},{0,1},{1,0},{0,-1}};queue<Pos> q;Pos v;while (cin >> n >> m) {for (i = 0; i < n; i++)for (j = 0; j < m; j++) {cin >> M[i][j];if (M[i][j] == 'r') {bi = i; bj = j;}visited[i][j] = false;if (M[i][j] == 'x') x[i][j] = true;}while (!q.empty())//清空上次监狱地图q.pop();flag = false;v.i = bi; v.j = bj; v.time = 0;q.push(v);visited[bi][bj] = true;//标记初始点已访问while (!q.empty()) {v = q.front();q.pop();i = v.i;j = v.j;time = v.time;if (M[i][j] == 'a') {flag = true; break;}//到达目的地if (M[i][j] == 'x' && x[i][j]) {//首次遇到该守卫,x[i][j] = false;v.time = time+1;//再次将他入队，因为他需要多消耗一个时间q.push(v);continue;}int k;for (k = 0; k < 4; k++) { //访问下一层点if (M[i + dir[k][0]][j + dir[k][1]] != '#' && !visited[i + dir[k][0]][j + dir[k][1]] && (i + dir[k][0]) >= 0 && (j + dir[k][1]) >= 0 && (i + dir[k][0]) < n && (j + dir[k][1]) < m ) {if (M[i + dir[k][0]][j + dir[k][0]] == 'x')//如果是守卫的点x[i + dir[k][0]][j + dir[k][1]] = true;v.i = i + dir[k][0]; v.j = j + dir[k][1]; v.time = time+1;//将当前点入队q.push(v);visited[i + dir[k][0]][j + dir[k][1]] = true;//标记已访问}}}if (flag)cout << time << endl;elsecout << "Poor ANGEL has to stay in the prison all his life." << endl;}return 0;}