UVa——1600（巡逻机器人）

迷宫求最短路的话一般用bfs都可以解决，但是这个加了个状态，那么就增加一个维度，用来判断k的值。比较简单的三维bfs。写搜索题的话一定要注意细节。这个题花了好长的时间。因为k的原因，一开始用了k的原因，dfs好想一些，因为可以回溯，k的值搜完一个方向，然后回溯。那样就很简单。而且数据是20*20.

但是最后dfs还是T了。

AC的bfs

 #include <iostream>
 #include <cstring>
 #include <string>
 #include <map>
 #include <set>
 #include <algorithm>
 #include <fstream>
 #include <cstdio>
 #include <cmath>
 #include <stack>
 #include <queue>
 using namespace std;
 const double Pi=3.14159265358979323846;
 typedef long long ll;
 const int MAXN=+;
 int dx[]={-,,,};
 int dy[]={,,,-};
 const int INF = 0x3f3f3f3f;
 const int NINF = 0xc0c0c0c0;
 const ll mod=1e9+;
 int vis[][][];
 int M[][];
 int m,n;
 struct node{
     int x,y,step,k;
     node (int x=,int y=,int step=,int k=)
     {
         this->x=x;
         this->y=y;
         this->step=step;
         this->k=k;
     }
 }tim,now;
 int bfs(int x,int y,int k)
 {
     queue <node> Q;
     Q.push(node(,,,k));
     vis[][][k]=;
     while(!Q.empty())
     {
         tim=Q.front();Q.pop();
         if(tim.x==m&&tim.y==n)
         {
             return tim.step;
         }
         for(int i=;i<;i++)
         {
             now.x=dx[i]+tim.x;
             now.y=dy[i]+tim.y;
             if(M[now.x][now.y]==)    now.k=tim.k-;
                     else now.k=k;
             if(now.x>=&&now.x<=m&&now.y>=&&now.y<=n&&(M[now.x][now.y]==||now.k>=)&&!vis[now.x][now.y][now.k])
             {
                 now.step=tim.step+;
                 Q.push(now);
                 //cout << now.x<<" "<< now.y<< " "<< now.step<<endl;
                 vis[now.x][now.y][now.k]=;
             }
         }
     }
     return -;
 }
 int main()
 {
     int t;cin>>t;
     while(t--)
     {
         cin>>m>>n;
         int k;cin>>k;
         memset(vis,,sizeof(vis));
         for(int i=;i<=m;i++)
             for(int j=;j<=n;j++)
                 cin>>M[i][j];
         cout <<bfs(,,k)<<endl;
     }
     return ;
 }

TLE的dfs

 #include <iostream>
 #include <cstring>
 #include <string>
 #include <map>
 #include <set>
 #include <algorithm>
 #include <fstream>
 #include <cstdio>
 #include <cmath>
 #include <stack>
 #include <queue>
 using namespace std;
 const double Pi=3.14159265358979323846;
 typedef long long ll;
 const int MAXN=+;
 int dx[]={-,,,};
 int dy[]={,,,-};
 const int INF = 0x3f3f3f3f;
 const int NINF = 0xc0c0c0c0;
 const ll mod=1e9+;
 int m,n;
 int vis[][];
 int M[][];
 int step[][];
 int ans=INF;
 int tk;
 void dfs(int x,int y,int k)
 {
     if(x==m&&y==n)
     {
         if(ans>step[x][y])
             ans=step[x][y];
         return;
     }
     for(int i=;i<;i++)
     {
         int sx=dx[i]+x;
         int sy=dy[i]+y;
         if(sx>=&&sx<=m&&sy>=&&sy<=n&&!vis[sx][sy]&&(k>||M[sx][sy]==))
         {
             int pk=k;
             if(M[sx][sy]==) pk--;
                 else pk=tk;
             vis[sx][sy]=;
             step[sx][sy]=step[x][y]+;
             dfs(sx,sy,pk);
             vis[sx][sy]=;
         }
     }
 }
 int main()
 {
     int t;cin>>t;
     while(t--)
     {
         ans=INF;
         memset(vis,,sizeof(vis));
         memset(step,-,sizeof(step));
         cin>>m>>n;cin>>tk;
         for(int i=;i<=m;i++)
             for(int j=;j<=n;j++)
                 cin>>M[i][j];
         step[][]=;
         vis[][]=;
         dfs(,,tk);
         if(step[m][n]!=-) cout <<ans<<endl;
             else cout <<-<<endl;
     }
     return ;
 }