BFS+模拟 ZOJ 3865 Superbot

 

题目传送门

 1 /*  2  BFS+模拟：dp[i][j][p] 表示走到i，j，方向为p的步数为多少；  3  BFS分4种情况入队，最后在终点4个方向寻找最小值:)  4 */  5 #include <cstdio>  6 #include <iostream>  7 #include <algorithm>  8 #include <cstring>  9 #include <string>  10 #include <queue>  11 using namespace std;  12  13 const int MAXN = 1e2 + 10;  14 const int INF = 0x3f3f3f3f;  15 int dir[4][4][2] = {  16 {{0, -1}, {0, 1}, {-1, 0}, {1, 0}},  17 {{1, 0}, {0, -1}, {0, 1}, {-1, 0}},  18 {{-1, 0}, {1, 0}, {0, -1}, {0, 1}},  19 {{0, 1}, {-1, 0}, {1, 0}, {0, -1}}  20 };  21 int dp[12][12][4];  22 int dp2[12][12][4];  23 int ans[12][12][4];  24 char maze[12][12];  25 int sx, sy, ex, ey;  26 int n, m, P;  27  28 bool ok(int nx, int ny, int np)  29 {  30 if (nx < 0 || nx >= n || ny < 0 || ny >= m) return false;  31 if (maze[nx][ny] == '*' || dp[nx][ny][np] != -1) return false;  32  33 return true;  34 }  35  36 void BFS(void)  37 {  38 int x, y, nx, ny, p, np;  39 queue<int> q;  40 q.push (sx); q.push (sy); q.push (0);  41  42 while (!q.empty ())  43  {  44 x = q.front (); q.pop ();  45 y = q.front (); q.pop ();  46 p = q.front (); q.pop ();  47  48 if (x == ex && y == ey && ans[x][y][p] == -1)  49 ans[x][y][p] = dp[x][y][p];  50  51 nx = x + dir[dp[x][y][p]/P%4][p][0];  52 ny = y + dir[dp[x][y][p]/P%4][p][1];  53 if (ok (nx, ny, p)) //move robot, not move dir  54  {  55 dp[nx][ny][p] = dp[x][y][p] + 1;  56  q.push (nx); q.push (ny); q.push (p);  57  }  58  59 np = p + 1;  60 if (np > 3) np = 0;  61 if (ok (x, y, np)) //not move robot, move dir to right  62  {  63 dp[x][y][np] = dp[x][y][p] + 1;  64  q.push (x); q.push (y); q.push (np);  65  }  66  67 np = p - 1;  68 if (np < 0) np = 3;  69 if (ok (x, y, np)) //not move robot, move dir to left  70  {  71 dp[x][y][np] = dp[x][y][p] + 1;  72  q.push (x); q.push (y); q.push (np);  73  }  74  75 if (dp[x][y][p] <= 200) //not move  76  {  77 dp[x][y][p]++;  78  q.push (x); q.push (y); q.push (p);  79  }  80  }  81  82 return ;  83 }  84  85 int main(void) //ZOJ 3865 Superbot  86 {  87 //freopen ("F.in", "r", stdin);  88  89 int t;  90 scanf ("%d", &t);  91 while (t--)  92  {  93 memset (dp, -1, sizeof (dp));  94 memset (ans, -1, sizeof (ans));  95  96 scanf ("%d%d%d", &n, &m, &P);  97  98 for (int i=0; i<n; ++i) scanf ("%s", &maze[i]);  99 for (int i=0; i<n; ++i) 100  { 101 for (int j=0; j<m; ++j) 102  { 103 if (maze[i][j] == '@') sx = i, sy = j; 104 else if (maze[i][j] == '$') ex = i, ey = j; 105  } 106  } 107 108 dp[sx][sy][0] = 0; 109  BFS (); 110 111 int mn = INF; 112 for (int i=0; i<4; ++i) 113  { 114 if (ans[ex][ey][i] == -1) continue; 115 mn = min (mn, ans[ex][ey][i]); 116  } 117 118 if (mn == INF) printf ("YouBadbad\n"); 119 else printf ("%d\n", mn); 120  } 121 122 return 0; 123 }