1 /*
2 题意：国际象棋起点到终点移动的最少步数
3
4 题解：水题，BFS就过了，有人用双向BFS
5 */
6 #include <cstdio>
7 #include <cstring>
8 #include <queue>
9
10 using namespace std;
11
12 int dir[8][2]={-2,1,-1,2,1,2,2,1,2,-1,1,-2,-1,-2,-2,-1};
13
14 bool gra[305][305];
15
16 struct node
17 {
18 int x,y,step;
19 bool operator==(const node & t)const
20 {
21 if (x == t.x && y == t.y)
22 return true;
23 else
24 return false;
25 }
26 }Q[90005];
27 int front,tail;
28 int bfs(int l, node start, node end)
29 {
30 front = tail = 0;
31 Q[tail++] = start;
32 gra[start.x][start.y] = true;
33 while (front < tail)
34 {
35 node now = Q[front++];
36 if (now == end)
37 return now.step;
38 for(int i=0; i<8; i++)
39 {
40 int nx = now.x + dir[i][0];
41 int ny = now.y + dir[i][1];
42 if (0 <= nx && nx < l && 0 <= ny && ny < l && !gra[nx][ny])
43 {
44 node tmp = now;
45 tmp.step = now.step + 1;
46 tmp.x = nx;
47 tmp.y = ny;
48 Q[tail++] = tmp;
49 gra[nx][ny] = true;
50 }
51 }
52 }
53 }
54
55 int main(void)
56 {
57 int t;
58 scanf("%d",&t);
59 while (t--)
60 {
61 int l;
62 scanf("%d",&l);
63 node start;
64 node end;
65 scanf("%d%d%d%d",&start.x,&start.y,&end.x,&end.y);
66 start.step = 0;
67 memset(gra,false,sizeof(gra));
68 printf("%d\n",bfs(l,start,end));
69 }
70 return 0;
71 }
双向BFS实现：
1 /*
2 题意：国际象棋起点到终点移动的最少步数
3
4 题解：参考网上其它人的代码写了个双向bfs，也许这个双向用得不太好，没快多少
5 */
6 #include <cstdio>
7 #include <cstring>
8 #include <queue>
9
10 #define FOR 2
11 #define BACK 3
12
13 using namespace std;
14
15 int dir[8][2]={-2,1,-1,2,1,2,2,1,2,-1,1,-2,-1,-2,-2,-1};
16
17 int gra[305][305];
18 int steps[305][305];
19
20 struct node
21 {
22 int x,y;
23 bool operator==(const node & t)const
24 {
25 if (x == t.x && y == t.y)
26 return true;
27 else
28 return false;
29 }
30 }fQ[45005],bQ[45005];
31 int ffront,ftail,bfront,btail;
32 int bi_bfs(int l, node start, node end)
33 {
34 if (start == end)
35 return 0;
36 ffront = ftail = bfront = btail = 0;
37 fQ[ftail++] = start;
38 bQ[btail++] = end;
39 gra[start.x][start.y] = FOR;
40 gra[end.x][end.y] = BACK;
41 while (ffront < ftail && bfront < btail)
42 {
43 if (ftail - ffront <= btail - bfront)
44 {
45 node now = fQ[ffront++];
46 for(int i=0; i<8; i++)
47 {
48 int nx = now.x + dir[i][0];
49 int ny = now.y + dir[i][1];
50 if (0 <= nx && nx < l && 0 <= ny && ny < l)
51 {
52 if (!gra[nx][ny])
53 {
54 node tmp;
55 tmp.x = nx;
56 tmp.y = ny;
57 steps[nx][ny] = steps[now.x][now.y] + 1;
58 fQ[ftail++] = tmp;
59 gra[nx][ny] = FOR;
60 }
61 else if (gra[nx][ny] == BACK)
62 return steps[nx][ny] + steps[now.x][now.y] + 1;
63 }
64 }
65 }
66 else
67 {
68 node now = bQ[bfront++];
69 for(int i=0; i<8; i++)
70 {
71 int nx = now.x + dir[i][0];
72 int ny = now.y + dir[i][1];
73 if (0 <= nx && nx < l && 0 <= ny && ny < l)
74 {
75 if (!gra[nx][ny])
76 {
77 node tmp;
78 tmp.x = nx;
79 tmp.y = ny;
80 steps[nx][ny] = steps[now.x][now.y] + 1;
81 bQ[btail++] = tmp;
82 gra[nx][ny] = BACK;
83 }
84 else if (gra[nx][ny] == FOR)
85 return steps[nx][ny] + steps[now.x][now.y] + 1;
86 }
87 }
88 }
89 }
90 }
91
92 int main(void)
93 {
94 int t;
95 scanf("%d",&t);
96 while (t--)
97 {
98 int l;
99 scanf("%d",&l);
100 node start;
101 node end;
102 scanf("%d%d%d%d",&start.x,&start.y,&end.x,&end.y);
103 memset(steps,0,sizeof(steps));
104 memset(gra,0,sizeof(gra));
105 printf("%d\n",bi_bfs(l,start,end));
106 }
107 return 0;
108 }
时间： 2024-10-20 13:06:22