POJ 2251 *** Dungeon Master

来源：互联网发布：软件体系结构风格编辑：程序博客网时间：2024/06/15 08:29

题意:有一个R*C*L的三维数组，从S走到E点，其中‘#’点不可到达，‘.’可到达。如果能够到达E点，那么最少需要多少步，如果不能输出不能。

想法:简单bfs，一开始用dfsTLE了，同时发现：dfs(position m,int step){},其中position为struct position{int x,int y,int z;}，poj的判题系统对于dfs({1,1,1},5)是编译不能通过的。。。

代码如下:

#pragma warning(disable:4996)#include<iostream>#include<cstdio>#include<cmath>#include<stack>#include<queue>#include<cstring>#include<sstream>#include<set>#include<string>#include<iterator>#include<vector>#include<map>#include<algorithm>using namespace std;const int INF = 0x7fffffff;struct position{int x, y, z;};int dir[6][3] = { {0,0,1},{0,0,-1},{0,1,0},{0,-1,0},{1,0,0},{-1,0,0} };int L, R, C, cnt;position start, goal;char maze[35][35][35];int mark[35][35][35];void bfs() {queue<position>path;path.push(start);int flag = 1;while (!path.empty()&&flag) {position now = path.front();path.pop();for (int d = 0; d < 6; ++d) {position temp;temp.x = now.x + dir[d][0], temp.y = now.y + dir[d][1], temp.z = now.z + dir[d][2];if (maze[temp.x][temp.y][temp.z] == 'E') {mark[temp.x][temp.y][temp.z] = mark[now.x][now.y][now.z]+1;flag = 0;break;}if (temp.x > 0 && temp.y > 0 && temp.z > 0 && temp.x <= L&&temp.y <= R&&temp.z <= C&&maze[temp.x][temp.y][temp.z] != '#'&&!mark[temp.x][temp.y][temp.z]) {mark[temp.x][temp.y][temp.z] = mark[now.x][now.y][now.z] + 1;path.push(temp);}}}}int main(void) {//freopen("input.txt", "r", stdin);//freopen("output.txt", "w", stdout);while (cin >> L >> R >> C, L || R || C) {for (int i = 1; i <= L; ++i)for (int j = 1; j <= R; ++j)for (int k = 1; k <= C; ++k) {cin >> maze[i][j][k];if (maze[i][j][k] == 'S') { start.x = i, start.y = j, start.z = k;}else if (maze[i][j][k] == 'E') { goal.x = i, goal.y = j, goal.z = k;}}memset(mark, 0, sizeof(mark));bfs();if (mark[goal.x][goal.y][goal.z]== 0)cout << "Trapped!" << endl;else cout << "Escaped in " << mark[goal.x][goal.y][goal.z] << " minute(s)." << endl;}}

0 0