UVA 816 Abbott's Revenge

发表于 | 分类于 ，

题意：迷宫中的点从不同方向进入时有不同的转向，求起点到终点的最短路。

每个点除了坐标x，y外，还有方向属性d。用BFS求最短路，存储BFS树中结点的父结点用于路径还原。

学习了一种用strchr来得到字符id的写法。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
#include <bits/stdc++.h>

using namespace std;

typedef pair<int, int> Pair;

const int MAX_N = 9 + 3;

int sx, sy, tx, ty;
string sd;
int maze[MAX_N][MAX_N][4];
vector<Pair> ans;

int dX[4] = {1, 0, -1, 0}, dY[4] = {0, -1, 0, 1};

struct Point { int x, y, d; };
Point pre[MAX_N][MAX_N][4];

const char* dir = "SWNE";
const char* adir = "LFR";

int did(char c) { return strchr(dir, c) - dir; }
int adid(char c) { return strchr(adir, c) - adir; }

bool bfs() {
  queue<Point> que;
  int x, y, d, fd = did(sd[0]);
  int nx = sx + dX[fd], ny = sy + dY[fd];
  pre[nx][ny][fd] = (Point) {sx, sy, fd};
  que.push((Point) {nx, ny, fd});

  while (!que.empty()) {
    Point p = que.front(); que.pop();
    x = p.x, y = p.y, d = p.d;
    if (x == tx && y == ty) break;

    for (int i = 0; i < 3; i++) {
      if (!((maze[x][y][d] >> i) & 1)) continue;
      fd = d;
      if (i == 0) fd = (fd - 1 + 4) % 4;
      else if (i == 2) fd = (fd + 1) % 4;

      int nx = x + dX[fd], ny = y + dY[fd];
      Point& pp = pre[nx][ny][fd];
      if (pp.x == 0 && pp.y == 0 && pp.d == 0) {
        pp = (Point) {x, y, d};
        que.push((Point) {nx, ny, fd});
      }
    }
  }

  for (int i = 0; i < 4; i++) {
    Point pp = pre[tx][ty][i];
    if (pp.x == 0 && pp.y == 0 && pp.d == 0) continue;
    Point p = (Point) {tx, ty, i};
    while (p.x != sx || p.y != sy) {
      ans.push_back(Pair(p.x, p.y));
      p = pre[p.x][p.y][p.d];
    }
    ans.push_back(Pair(sx, sy));
    reverse(ans.begin(), ans.end());
    return true;
  }
  return false;
}

int main() {
  string name, ad;
  int x, y;
  while (cin >> name && name != "END") {
    memset(maze, 0, sizeof(maze));
    memset(pre, 0, sizeof(pre));
    ans.clear();

    cin >> sx >> sy >> sd >> tx >> ty;
    while (cin >> x && x) {
      cin >> y;
      while (cin >> ad && ad != "*") {
        for (int i = 1; i < ad.size(); i++) {
          maze[x][y][did(ad[0])] |= 1 << adid(ad[i]);
        }
      }
    }

    bool res = bfs();
    cout << name << endl;
    if (res) {
      for (int i = 0; i < ans.size(); i++) {
        if (i % 10 == 0) cout << "  ";
        cout << "(" << ans[i].first << "," << ans[i].second << ")";
        if (i != ans.size() - 1)
          cout << ((i + 1) % 10 == 0 ? '\n' : ' ');
      }
      cout << endl;
    } else {
      cout << "  No Solution Possible" << endl;
    }
  }
  return 0;
}