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 = ;
int n, m;

char str[][maxn][maxn];

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

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

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