BFS+贪心 HDOJ 5335 Walk Out

 

题目传送门

 1 /*
 2     题意：求从(1, 1)走到(n, m)的二进制路径值最小
 3     BFS+贪心：按照标程的作法，首先BFS搜索所有相邻0的位置，直到1出现。接下去从最靠近终点的1开始，
 4             每一次走一步，不走回头路，只往下或往右走。因为满足i = j + (i - j)的坐标(j, i - j)可能不止一个，选择最小的访问
 5 */
 6 #include 
 7 #include 
 8 #include 
 9 #include 
10 #include 
11 #include 
12 using namespace std;
13 
14 const int MAXN = 1e3 + 10;
15 const int INF = 0x3f3f3f3f;
16 int dx[4] = {0, 1, 0, -1};
17 int dy[4] = {1, 0, -1, 0};
18 char maze[MAXN][MAXN];
19 bool vis[MAXN][MAXN];
20 int n, m;
21 
22 bool judge(int x, int y)    {
23     if (x < 1 || x > n || y < 1 || y > m || vis[x][y]) return false;
24     return true;
25 }
26 
27 void BFS(void)  {
28     memset (vis, false, sizeof (vis));  int mx = 2;
29     queueint, int> > Q;   Q.push (make_pair (1, 1));  vis[1][1] = true;
30     while (!Q.empty ()) {
31         int x = Q.front ().first, y = Q.front ().second; Q.pop ();
32         if (maze[x][y] == '0')  {
33             for (int i=0; i<4; ++i) {
34                 int tx = x + dx[i], ty = y + dy[i];
35                 if (!judge (tx, ty))    continue;
36                 mx = max (mx, tx + ty);
37                 Q.push(make_pair (tx, ty));    vis[tx][ty] = true;
38             }
39         }
40     }
41 
42     if (vis[n][m] && maze[n][m] == '0') {
43         puts ("0"); return ;
44     }
45     
46     printf ("1");
47     for (int i=mx; ii) {
48         char mn = '1';
49         for (int j=1; j<=n; ++j)    {
50             if (1 <= i - j && i - j <= m && vis[j][i-j])    {
51                 mn = min (mn, maze[j+1][i-j]);
52                 mn = min (mn, maze[j][i-j+1]);
53             }
54         }
55         printf ("%c", mn);
56         for (int j=1; j<=n; ++j)    {
57             if (1 <= i - j && i - j <= m && vis[j][i-j])    {
58                 if (maze[j+1][i-j] == mn) vis[j+1][i-j] = true;
59                 if (maze[j][i-j+1] == mn) vis[j][i-j+1] = true;
60             }
61         }
62     }
63     puts ("");
64 }
65 
66 int main(void)  {       //HDOJ 5335 Walk Out
67     //freopen ("I.in", "r", stdin);
68     int T;  scanf ("%d", &T);
69     while (T--) {
70         scanf ("%d%d", &n, &m);
71         for (int i=1; i<=n; ++i) scanf ("%s", maze[i] + 1);
72         for (int i=1; i<=n; ++i)  maze[i][0] = maze[i][m+1] = '2';
73         for (int i=0; i<=m+1; ++i)  maze[0][i] = maze[n+1][i] = '2';
74         BFS ();
75     }
76 
77     return 0;
78 }

 

转载于:https://www.cnblogs.com/Running-Time/p/4691182.html