离散化+BFS HDOJ 4444 Walk

转载

mob60475702c725 2015-07-18 14:05:00

1 /*
  2     题意：问一个点到另一个点的最少转向次数。
  3     坐标离散化+BFS：因为数据很大，先对坐标离散化后，三维(有方向的)BFS
  4         关键理解坐标离散化，BFS部分可参考HDOJ_1728
  5 */
  6 #include <cstdio>
  7 #include <algorithm>
  8 #include <cstring>
  9 #include <queue>
 10 #include <vector>
 11 #include <map>
 12 #include <cmath>
 13 using namespace std;
 14 
 15 const int MAXN = 8 * 50 + 100;
 16 const int INF = 0x3f3f3f3f;
 17 vector<int> nx, ny;
 18 map<int, int> mx, my;
 19 struct Point
 20 {
 21     int x, y, t, d;
 22     Point (int _x = 0, int _y = 0)    {x = _x, y = _y, t = 0;}
 23     void read(void)
 24     {
 25         scanf ("%d%d", &x, &y);
 26         nx.push_back (x);    ny.push_back (y);
 27     }
 28     void updata(void) {x = mx[x];    y = my[y];}
 29     bool operator < (const Point &r) const    {return t > r.t;}
 30 }s, e, p[55][4];
 31 bool maze[2*MAXN][2*MAXN];
 32 int dp[MAXN][MAXN][4];
 33 bool can[MAXN][MAXN][4];
 34 int dx[4] = {0, 1, 0, -1};
 35 int dy[4] = {1, 0, -1, 0};
 36 int w;
 37 
 38 int compress(vector<int> &x, map<int, int> &mp)
 39 {
 40     vector<int> xs;
 41     sort (x.begin (), x.end ());
 42     x.erase (unique (x.begin (), x.end ()), x.end ());
 43     for (int i=0; i<x.size (); ++i)
 44     {
 45         for (int d=-1; d<=1; ++d)    xs.push_back (x[i] + d);
 46     }
 47     sort (xs.begin (), xs.end ());
 48     xs.erase (unique (xs.begin (), xs.end ()), xs.end ());
 49     for (int i=0; i<xs.size (); ++i)    mp[x[i]] = find (xs.begin (), xs.end (), x[i]) - xs.begin ();
 50     return xs.size ();
 51 }
 52 
 53 bool check(Point &a)
 54 {
 55     if (0 <= a.x && a.x <= w && 0 <= a.y && a.y <= w && dp[a.x][a.y][a.d] > a.t)
 56     {
 57         dp[a.x][a.y][a.d] = a.t;
 58         return true;
 59     }
 60     return false;
 61 }
 62 
 63 int BFS(void)
 64 {
 65     memset (dp, INF, sizeof (dp));
 66     priority_queue<Point> Q;    s.t = 0;
 67     for (s.d=0; s.d<4; ++s.d)
 68     {
 69         Q.push (s);    dp[s.x][s.y][s.d] = 0;
 70     }
 71 
 72     while (!Q.empty ())
 73     {
 74         Point now = Q.top ();    Q.pop ();
 75         if (dp[now.x][now.y][now.d] < now.t)    continue;
 76         if (now.x == e.x && now.y == e.y)    return now.t;
 77         for (int d=-1; d<=1; ++d)
 78         {
 79             Point to = now;
 80             to.d = (to.d + 4 + d) % 4;
 81             if (!can[to.x][to.y][to.d])    continue;
 82             int x = 2 * to.x, y = 2 * to.y;
 83             if (maze[x][y] && maze[x+1][y+1] &&
 84                 ((now.d % 2 == 0 && d != 1) || (now.d % 2 == 1 && d != -1)))    continue;
 85             if (maze[x+1][y] && maze[x][y+1] &&
 86                 ((now.d % 2 == 0 && d != -1) || (now.d % 2 == 1 && d != 1)))    continue;
 87             if (d != 0)    to.t++;
 88             to.x += dx[to.d];    to.y += dy[to.d];
 89             if (check (to))    Q.push (to);
 90         }
 91     }
 92 
 93     return -1;
 94 }
 95 
 96 int main(void)        //HDOJ 4444 Walk
 97 {
 98 //    freopen ("C.in", "r", stdin);
 99 
100     while (true)
101     {
102         nx.clear ();    ny.clear ();    mx.clear ();    my.clear ();
103         s.read ();    e.read ();
104         if (!s.x && !s.y && !e.x && !e.y)    break;
105         memset (maze, false, sizeof (maze));
106 
107         int n;    scanf ("%d", &n);
108         for (int i=1; i<=n; ++i)
109         {
110             Point *t = p[i];
111             t[0].read ();    t[2].read ();
112             if (t[0].x > t[2].x)    swap (t[0], t[2]);
113             if (t[0].y > t[2].y)
114             {
115                 Point a = t[0], b = t[2];
116                 t[0].y = b.y;    t[2].y = a.y;
117             }
118             t[1] = (Point) {t[2].x, t[0].y};
119             t[3] = (Point) {t[0].x, t[2].y};
120         }
121 
122         w = max (compress (nx, mx), compress (ny, my));
123         s.updata ();    e.updata ();
124         for (int i=1; i<=n; ++i)
125         {
126             Point *t = p[i];
127             for (int j=0; j<4; ++j)    t[j].updata ();
128             for (int j=0; j<4; ++j)    t[j] = Point (2*t[j].x, 2*t[j].y);
129             for (int j=t[0].x+1; j<=t[2].x; ++j)
130             {
131                 for (int k=t[0].y+1; k<=t[2].y; ++k)    maze[j][k] = true;        //离散化后将矩形涂黑
132             }
133         }
134 
135         memset (can, true, sizeof (can));
136         for (int i=0; i<w; ++i)
137         {
138             for (int j=0; j<w; ++j)
139             {
140                 int x = i * 2, y = j * 2;
141                 bool *d = can[i][j];
142                 if (maze[x][y+1] && maze[x+1][y+1])    d[0] = false;        //判断4个方向能不能走
143                 if (maze[x+1][y] && maze[x+1][y+1])    d[1] = false;
144                 if (maze[x][y] && maze[x+1][y])    d[2] = false;
145                 if (maze[x][y] && maze[x][y+1])    d[3] = false;
146             }
147         }
148 
149         printf ("%d\n", BFS ());
150     }
151 
152     return 0;
153 }