AtCoder Beginner Contest 132 E - Hopscotch Addict

bfs

位置+状态

just need to calculate min value(only it is useful), so O(1*x)

挺有趣的一道题。。。

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cmath>
 4 #include <cstring>
 5 #include <string>
 6 #include <algorithm>
 7 #include <iostream>
 8 using namespace std;
 9 #define ll long long
10
11 const int maxn=1e6+10;
12
13 struct node
14 {
15     int d;
16     node *to;
17 }*e[maxn];
18
19 int qx[maxn*3],qy[maxn*3],qz[maxn*3];
20
21 int f[maxn][3];
22
23 int main()
24 {
25     node *p;
26     int n,m,x,y,z,xx,yy,zz,head,tail,s,t;
27     scanf("%d%d",&n,&m);
28     while (m--)
29     {
30         scanf("%d%d",&x,&y);
31         p=new node();
32         p->d=y;
33         p->to=e[x];
34         e[x]=p;
35     }
36
37     scanf("%d%d",&s,&t);
38     qx[1]=s,qy[1]=0,qz[1]=3;
39     head=0,tail=1;
40     while (head<tail)
41     {
42         head++;
43         x=qx[head];
44         y=qy[head];
45         z=qz[head];
46
47         p=e[x];
48         while (p)
49         {
50             xx=p->d;
51             yy=(y+1)%3;
52             zz=z+1;
53             if (!f[xx][yy])
54             {
55                 if (xx==t && yy==0)
56                 {
57                     printf("%d",(zz-3)/3);
58                     return 0;
59                 }
60                 tail++;
61                 qx[tail]=xx;
62                 qy[tail]=yy;
63                 qz[tail]=zz;
64                 f[xx][yy]=zz;
65             }
66             p=p->to;
67         }
68     }
69     printf("-1");
70     return 0;
71 }

原文地址：https://www.cnblogs.com/cmyg/p/11108069.html