bzoj 3299 [USACO2011 Open]Corn Maze玉米迷宫（最短路）

Posted 2020-08-17 konjak魔芋

 

就一个最短路，并且边长都是1，所以每个点只搜一次。

 

 1 /**************************************************************
 2     Problem: 3299
 3     User: MT_Chan
 4     Language: C++
 5     Result: Accepted
 6     Time:72 ms
 7     Memory:2420 kb
 8 ****************************************************************/
 9  
10 #include<cstdio>
11 #include<cstdlib>
12 #include<cstring>
13 #include<iostream>
14 #include<algorithm>
15 #include<queue>
16 using namespace std;
17 #define Maxn 310
18 #define INF 0xfffffff
19  
20 int kx[30];
21 int a[Maxn][Maxn],tr[Maxn*Maxn];
22 char s[Maxn];
23  
24 int dis[Maxn*Maxn],st,ed;
25 int n,m;
26  
27 queue<int > q;
28 void spfa()
29 {
30     while(!q.empty()) q.pop();
31     memset(dis,-1,sizeof(dis));
32     q.push(st);dis[st]=0;
33     while(!q.empty())
34     {
35         int x=q.front();
36         int nx=(x-1)/m+1,ny=x-(nx-1)*m;
37         if(ny>1&&tr[x-1]!=-1&&dis[tr[x-1]]==-1)
38         {
39             dis[tr[x-1]]=dis[x]+1;
40             if(tr[x-1]==ed) break;
41             q.push(tr[x-1]);
42         }
43         if(ny<m&&tr[x+1]!=-1&&dis[tr[x+1]]==-1)
44         {
45             dis[tr[x+1]]=dis[x]+1;
46             if(tr[x+1]==ed) break;
47             q.push(tr[x+1]);
48         }
49         if(nx>1&&tr[x-m]!=-1&&dis[tr[x-m]]==-1)
50         {
51             dis[tr[x-m]]=dis[x]+1;
52             if(tr[x-m]==ed) break;
53             q.push(tr[x-m]);
54         }
55         if(nx<n&&tr[x+m]!=-1&&dis[tr[x+m]]==-1)
56         {
57             dis[tr[x+m]]=dis[x]+1;
58             if(tr[x+m]==ed) break;
59             q.push(tr[x+m]);
60         }
61         q.pop();
62     }
63     printf("%d\\n",dis[ed]);
64 }
65  
66 int main()
67 {
68     scanf("%d%d",&n,&m);
69     memset(kx,-1,sizeof(kx));
70     memset(tr,0,sizeof(tr));
71     for(int i=1;i<=n;i++)
72     {
73         scanf("%s",s);
74         for(int j=0;j<m;j++)
75         {
76             int now=(i-1)*m+j+1;
77             if(s[j]==\'@\') st=now;
78             else if(s[j]==\'=\') ed=now;
79             else if(s[j]==\'#\') tr[now]=-1;
80             else if(s[j]>=\'A\'&&s[j]<=\'Z\')
81             {
82                 int kk=s[j]-\'A\'+1;
83                 if(kx[kk]==-1) kx[kk]=now;
84                 else tr[now]=kx[kk],tr[kx[kk]]=now;
85             }
86         }
87     }
88     for(int i=1;i<=n*m;i++) if(tr[i]==0) tr[i]=i;
89     spfa();
90     return 0; 
91 }

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