Codeforces Round #354 (Div. 2) D. Theseus and labyrinth

题目链接：

http://codeforces.com/contest/676/problem/D

题意：

如果两个相邻的格子都有对应朝向的门，则可以从一个格子到另一个格子，给你初始坐标xt，yt，终点坐标xm,ym，现在你可以选择在原地把地图上所有格子顺时针旋转90度；或者往上下左右走一格，问走到终点的最短路。

题解：

三维的bfs最短路，就是写起来比较麻烦。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<utility>
using namespace std;

const int maxn = 1111;
int n, m;

char str[4][maxn][maxn];

char mp[256],dir[256][4];
void get_mp(){
    memset(mp, 0, sizeof(mp));
    mp['+'] = '+';
    mp['-'] = '|';
    mp['|'] = '-';
    mp['^'] = '>';
    mp['>'] = 'v';
    mp['v'] = '<';
    mp['<'] = '^';
    mp['L'] = 'U';
    mp['U'] = 'R';
    mp['R'] = 'D';
    mp['D'] = 'L';
    mp['*'] = '*';

    memset(dir, 0, sizeof(dir));
    dir['+'][0] = dir['+'][1] = dir['+'][2] = dir['+'][3] = 1;
    dir['-'][1]=dir['-'][3]=1;
    dir['|'][0]=dir['|'][2]=1;
    dir['^'][0]=1;
    dir['>'][1]=1;
    dir['v'][2]=1;
    dir['<'][3]=1;
    dir['L'][0]=dir['L'][1]=dir['L'][2]=1;
    dir['U'][1]=dir['U'][2]=dir['U'][3]=1;
    dir['R'][0]=dir['R'][2]=dir['R'][3]=1;
    dir['D'][0] = dir['D'][1] = dir['D'][3] = 1;
}

struct Node {
    int s, x, y;
    Node(int s, int x, int y) :s(s), x(x), y(y) {}
    Node() {}
};

int xt, yt, xm, ym;
int d[4][maxn][maxn];
int vis[4][maxn][maxn];
int bfs() {
    memset(d, 0x3f, sizeof(d));
    memset(vis, 0, sizeof(vis));
    d[0][xt][yt] = 0;
    queue<Node> Q;
    Q.push(Node(0, xt, yt)); vis[0][xt][yt] = 1;
    while (!Q.empty()) {
        Node nd = Q.front(); Q.pop();
        int s = nd.s, x = nd.x, y = nd.y;
        if (x == xm&&y == ym) return d[s][x][y];
        int ss, xx, yy;
        ss = (s + 1) % 4, xx = x, yy = y;
        if (!vis[ss][xx][yy]) {
            d[ss][xx][yy] = d[s][x][y] + 1;
            Q.push(Node(ss, xx, yy));
            vis[ss][xx][yy] = 1;
        }
        ss = s, xx = x - 1, yy = y;
        if (!vis[ss][xx][yy]&&xx>=1&&dir[str[ss][xx][yy]][2]&&dir[str[s][x][y]][0]) {
            d[ss][xx][yy] = d[s][x][y] + 1;
            Q.push(Node(ss, xx, yy));
            vis[ss][xx][yy] = 1;
        }
        ss = s, xx = x + 1, yy = y;
        if (!vis[ss][xx][yy] && xx <=n && dir[str[ss][xx][yy]][0] && dir[str[s][x][y]][2]) {
            d[ss][xx][yy] = d[s][x][y] + 1;
            Q.push(Node(ss, xx, yy));
            vis[ss][xx][yy] = 1;
        }
        ss = s, xx = x, yy = y-1;
        if (!vis[ss][xx][yy] && yy >= 1 && dir[str[ss][xx][yy]][1] && dir[str[s][x][y]][3]) {
            d[ss][xx][yy] = d[s][x][y] + 1;
            Q.push(Node(ss, xx, yy));
            vis[ss][xx][yy] = 1;
        }
        ss = s, xx = x, yy = y + 1;
        if (!vis[ss][xx][yy] && yy <= m && dir[str[ss][xx][yy]][3] && dir[str[s][x][y]][1]) {
            d[ss][xx][yy] = d[s][x][y] + 1;
            Q.push(Node(ss, xx, yy));
            vis[ss][xx][yy] = 1;
        }
    }
    return -1;
}

int main() {
    get_mp();
    while (scanf("%d%d", &n, &m) == 2 && n) {
        for (int i = 1; i <= n; i++) scanf("%s", str[0][i]+1);
        for (int i = 1; i <= 3; i++) {
            for (int j = 1; j <= n; j++) {
                for (int k = 1; k <= m; k++) {
                    str[i][j][k] = mp[str[i - 1][j][k]];
                }
            }
        }
        scanf("%d%d%d%d", &xt, &yt, &xm, &ym);
        printf("%d
", bfs());
    }
    return 0;
}