USACO4.4 Shuttle Puzzle【bfs+优化】

直接上$bfs$,每一个状态记录下当前字符串的样子,空格的位置,和走到这个状态的答案。

用空格的位置转移,只有$50pts$

考虑到题目一个性质:$W$只往右走,$B$只往左走,就可以过了。

  1 #include<cstdio>
  2 #include<algorithm>
  3 #include<vector>
  4 #include<cstring>
  5 #include<queue>
  6 #include<map>
  7 #include<iostream>
  8 using namespace std;
  9 #define ll long long
 10 #define INF 0x3f3f3f3f
 11 struct node{
 12     int pos;
 13     string s,ans;
 14 };
 15 int n;
 16 queue<node>Q;
 17 map<string,bool>vis;
 18 string aim;
 19 string bfs()
 20 {
 21     node st;string aim="";
 22     st.s="";
 23     for(int i=1;i<=n;i++)
 24         st.s+='W',aim+='B';
 25     st.s+='@',aim+='@';
 26     for(int i=1;i<=n;i++)
 27         st.s+='B',aim+='W';
 28     st.pos=n;
 29     Q.push(st);
 30     vis[st.s]=1;
 31     while(!Q.empty())
 32     {
 33         node now=Q.front();Q.pop();
 34         if(now.s==aim)
 35             return now.ans;
 36         //cout<<now.s<<endl;
 37         int idx=now.pos,len=now.s.size();
 38         node nxt=now;
 39         if(idx-2>=0&&nxt.s[idx-2]!=nxt.s[idx-1]&&nxt.s[idx-2]=='W')
 40         {
 41             nxt.s[idx]=nxt.s[idx-2];
 42             nxt.s[idx-2]='@';
 43             //cout<<nxt.s<<endl;//
 44             char ch='0'+idx-2;
 45             nxt.ans=now.ans+ch;
 46             nxt.pos=idx-2;
 47             if(!vis[nxt.s])
 48                 Q.push(nxt);
 49             vis[nxt.s]=1;
 50         }
 51         nxt=now;
 52         if(idx-1>=0&&nxt.s[idx-1]=='W')
 53         {
 54             nxt.s[idx]=nxt.s[idx-1];
 55             nxt.s[idx-1]='@';
 56             //cout<<nxt.s<<endl;//
 57             char ch='0'+idx-1;
 58             nxt.ans=now.ans+ch;
 59             nxt.pos=idx-1;
 60             if(!vis[nxt.s])
 61                 Q.push(nxt);
 62             vis[nxt.s]=1;
 63         }
 64         nxt=now;
 65         if(idx+1<len&&nxt.s[idx+1]=='B')
 66         {
 67             nxt.s[idx]=nxt.s[idx+1];
 68             nxt.s[idx+1]='@';
 69             char ch='0'+idx+1;
 70             nxt.ans=now.ans+ch;
 71             nxt.pos=idx+1;
 72             if(!vis[nxt.s])
 73                 Q.push(nxt);
 74             vis[nxt.s]=1;
 75         }
 76         nxt=now;
 77         if(idx+2<len&&nxt.s[idx+2]!=nxt.s[idx+1]&&nxt.s[idx+2]=='B')
 78         {
 79             nxt.s[idx]=nxt.s[idx+2];
 80             nxt.s[idx+2]='@';
 81             char ch='0'+idx+2;
 82             nxt.ans=now.ans+ch;
 83             nxt.pos=idx+2;
 84             if(!vis[nxt.s])
 85                 Q.push(nxt);
 86             vis[nxt.s]=1;
 87         }
 88     }
 89     return "";
 90 }
 91 int main() 
 92 {
 93     //freopen("shuttle.in","r",stdin);
 94     //freopen("shuttle.out","w",stdout);
 95     scanf("%d",&n);
 96     string c=bfs();
 97     //cout<<c<<endl;
 98     int tot=0;
 99     for(int i=0;i<c.size();i++)
100     {    
101         tot++;
102         printf("%d",c[i]-'0'+1);
103         if(tot==20||i==c.size()-1)
104         {
105             puts("");
106             tot=0;
107         }
108         else printf(" ");
109     }
110     return 0;
111 }
Code
原文地址:https://www.cnblogs.com/lyttt/p/11846641.html