Poj 1915 - Knight Moves 双向广搜

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 
 4 int vis[305][305], mat[305][305];
 5 int dx[] = {-2, -2, -1, 1, 2, 2, 1, -1};
 6 int dy[] = {-1, 1, 2, 2, 1, -1, -2, -2};
 7 int casenum, nNum, sx, sy, tx, ty, i;
 8 
 9 struct point
10 {
11     int x, y;
12 }cur, next, q[90005]={0};
13 
14 int IsInBound(int x, int y)
15 {
16     return (x>=0 && y>=0 && x<nNum && y<nNum);
17 }/* IsInBound */
18 
19 int Solve()
20 {
21     int rear = -1;
22     int front = -1;
23     
24     cur.x = sx;
25     cur.y = sy;
26     vis[sx][sy] = 1; /* 从起始位置开始的探索标记为 1 */ 
27     q[++rear] = cur; /* 起始坐标入队 */ 
28     
29     next.x = tx;
30     next.y = ty;
31     vis[tx][ty] = 2;  /* 从终点位置开始的探索标记为 2 */ 
32     q[++rear] = next; /* 终点坐标入队 */ 
33     
34     while (front < rear)
35     {
36         cur = q[++front]; /* 队首节点坐标出队 */
37         
38         for (i=0; i<8; ++i)
39         {
40             next.x = cur.x + dx[i];
41             next.y = cur.y + dy[i];
42             
43             if (!IsInBound(next.x, next.y))
44                 continue;
45                 
46             if (!vis[next.x][next.y])
47             {
48                 vis[next.x][next.y] = vis[cur.x][cur.y];     /* 设为与当前探索路径相同的标记 */
49                 mat[next.x][next.y] = mat[cur.x][cur.y] + 1; /* 记录步数 */ 
50                 q[++rear] = next; /* 当前合法坐标位置入队 */ 
51             }
52             else if (vis[cur.x][cur.y] != vis[next.x][next.y])
53             {   /* 说明从起点出发的探索与从终点出发的探索重合 */ 
54                 return mat[cur.x][cur.y]+mat[next.x][next.y]+1;
55             }
56         }/* End of For */
57     }/* End of While */
58 }/* Solve */
59 
60 int main()
61 {
62     scanf("%d", &casenum);
63     while (casenum--)
64     {
65         memset(vis, 0, sizeof(vis));
66         memset(mat, 0, sizeof(mat));
67         
68         scanf("%d", &nNum);
69         scanf("%d %d", &sx, &sy);
70         scanf("%d %d", &tx, &ty);
71         
72         if (sx==tx && sy==ty)
73         {
74             printf("0\n");
75         }
76         else
77         {
78             printf("%d\n", Solve());
79         }    
80     }/* End of While */
81     
82     return 0;
83 }