离散化+BFS HDOJ 4444 Walk

 

题目传送门

 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 }