Jumping on Wils CodeForces, jumpingcodeforces

Jumping on Wils CodeForces, jumpingcodeforces

Jumping on Wils CodeForces-198B

It should be an Implicit Graph bfs or dp.

First, a tle o (nklogn)

 1 #include<cstdio> 2 #include<set> 3 using namespace std; 4 typedef pair<bool,int> P; 5 set<P> ss[2]; 6 P t; 7 char s[2][100100]; 8 int n,k,ii; 9 int main()10 {11     int i,num,hei;12     scanf("%d%d",&n,&k);13     scanf("%s",s[0]+1);14     scanf("%s",s[1]+1);15     if(s[0][1]!='X')    ss[0].insert(P(0,1));16     for(i=1;i<=n;i++)17     {18         ii^=1;19         ss[ii].clear();20         for(auto t:ss[ii^1])21         {22             num=t.first;23             hei=t.second;24             if(hei+k>n)25             {26                 puts("YES");27                 return 0;28             }29             if(hei-1>i&&s[num][hei-1]!='X')30                 ss[ii].insert(P(num,hei-1));31             if(hei+1>i&&s[num][hei+1]!='X')32                 ss[ii].insert(P(num,hei+1));33             if(hei+k>i&&s[num^1][hei+k]!='X')34                 ss[ii].insert(P(num^1,hei+k));35         }36     }37     puts("NO");38     return 0;39 }

Later, I realized the same position. It would not be better to go there at a later time after the earlier time. Therefore, I only need to traverse the same status once, we can optimize the complexity to O (nlogn) (isn't that obvious? Why didn't we think of it)

 1 #include<cstdio> 2 #include<set> 3 using namespace std; 4 typedef pair<bool,int> P; 5 set<P> ss[2]; 6 P t; 7 char s[2][100100]; 8 int n,k,ii; 9 bool vis[2][100100];10 int main()11 {12     int i,num,hei;13     scanf("%d%d",&n,&k);14     scanf("%s",s[0]+1);15     scanf("%s",s[1]+1);16     if(s[0][1]!='X')    ss[0].insert(P(0,1));17     for(i=1;i<=n;i++)18     {19         ii^=1;20         ss[ii].clear();21         for(auto t:ss[ii^1])22         {23             num=t.first;24             hei=t.second;25             if(hei+k>n)26             {27                 puts("YES");28                 return 0;29             }30             if(hei-1>i&&s[num][hei-1]!='X'&&(!vis[num][hei-1]))31                 ss[ii].insert(P(num,hei-1)),vis[num][hei-1]=1;32             if(hei+1>i&&s[num][hei+1]!='X'&&(!vis[num][hei+1]))33                 ss[ii].insert(P(num,hei+1)),vis[num][hei+1]=1;34             if(hei+k>i&&s[num^1][hei+k]!='X'&&(!vis[num^1][hei+k]))35                 ss[ii].insert(P(num^1,hei+k)),vis[num^1][hei+k]=1;36         }37     }38     puts("NO");39     return 0;40 }