TC SRM 637 DIV1 pathgame

Problem Statement

Give a 2*1000 board, each black or white. The white lattice has a common edge on the path,path adjacent to the first column and the last column from left to right. The two people took turns to paint the white lattice black lattice, the last one forced to black lattice so that the path disconnected people lose. Ask the winner to win or will fail.

(Translated by myq_952)

Tutorial

Do the problem of the SG is a bit raw ( there is nothing to live on DP it )

Consider a section of the 2*k of the whole white rectangle as a situation, blacked out a lattice will be the original situation into two successive situations, and then $n^2$ the situation of the SG can

It is important to note that there are 4 types of discussions (see Code) because the situation is legitimate and related to the export situation on both sides.

In addition, since there is no way to go as a must fail, then the value of the illegal situation of the SG is meaningless, it is important to ignore = =

While I was doing this, I called the other functions that also updated the time tag with time tags, causing the time stamp to be invalid. = This is Konjac Konjac

1#include <Set>2#include <ctime>3#include <queue>4#include <cstdio>5#include <bitset>6#include <cctype>7#include <bitset>8#include <cstdlib>9#include <cassert>Ten#include <cstring> One#include <iostream> A#include <algorithm> - #defineINF (1&LT;&LT;30) - #defineINF (1ll<<62) the #defineFi first - #defineSe Second - #defineRep (x,s,t) for (register int x=s,t_=t;x<t_;++x) - #definePer (x,s,t) for (register int x=t-1,s_=s;x>=s_;--x) + #defineTravel (x) for (int i=last[x],to;i&& (to=e[i].to); i=e[i].nxt) - #definePRT (x) cout<< #x << ":" <<x<< "" + #definePRTN (x) cout<< #x << ":" <<x<<endl A #definePB (x) push_back (x) at #defineHash asfmaljkg - #defineRank ASFJHGSKJF - #defineY1 Asggnja - #defineY2 SLFVM - using namespacestd; -typedefLong Longll; intypedef pair<int,int>II; -  to stringwin="Snuke"; + stringLose="sothe"; -  the Const intmaxn=1024x768; * intN; $ intMARK[MAXN],ALLC;Panax Notoginseng structsg_circle{ -     intsg[4][MAXN]; the     voidinit () { +memset (sg[0],-1, n+1<<2); Amemset (sg[1],-1, n+1<<2); thememset (sg[2],-1, n+1<<2); +memset (sg[3],-1, n+1<<2); -     } $     intSG0 (intx) { $         if(~sg[0][X])returnsg[0][x]; -         if(x<=1)returnsg[0][x]=x; -          theRep (I,0, X) SG1 (i); -         Wuyi++ALLC; theRep (I,0, x) MARK[SG1 (i) ^SG1 (x-i-1)]=ALLC; -Rep (I,0, MAXN)if(mark[i]!=ALLC) Wu             returnsg[0][x]=i; -     } About     //.... $     //.... -     intSG1 (intx) { -         if(~sg[1][X])returnsg[1][x]; -         if(x<=1)returnsg[1][x]=x; A          +Rep (I,0, X) SG1 (i), SG2 (i), SG3 (i); the          -++ALLC; $Rep (I,1, x+1){ the             if(x-i) MARK[SG1 (i-1) ^SG2 (x-i)]=ALLC; theMARK[SG1 (I-1) ^SG3 (x-i)]=ALLC; the         } theRep (I,0, MAXN)if(mark[i]!=ALLC) -             returnsg[1][x]=i; in     } the     //#... the     //.... About     intSG2 (intx) { the         if(~sg[2][X])returnsg[2][x]; the         if(x<=1)returnsg[2][x]=0;///no matter if you lose the          +Rep (I,0, X) SG2 (i), SG3 (i); -      the++ALLC;BayiRep (I,2, x+1) MARK[SG2 (I-1) ^SG3 (x-i)]=ALLC; theRep (I,0, MAXN)if(mark[i]!=ALLC) the             returnsg[2][x]=i; -     } -     //#... the     //...# the     intSG3 (intx) { the         if(~sg[3][X])returnsg[3][x]; the         if(x<=1)returnsg[3][x]=x; -          theRep (I,0, X) SG2 (i), SG3 (i); the      the++ALLC;94Rep (I,1, x+1){ the             if(I-1&&x-i) MARK[SG2 (i-1) ^SG2 (x-i)]=ALLC; theMARK[SG3 (I-1) ^SG3 (x-i)]=ALLC; the         }98Rep (I,0, MAXN)if(mark[i]!=ALLC) About             returnsg[3][x]=i; -     }101     //#.. #102     //....103 }SG;104 classPathgame { the  Public:106     stringJudge (Vector <string>d) {107n=d[0].length ();108 109 sg.init (); the         intans=0;111          for(intI=0, j;i<n;i++){ the              for(j=i;j<n&&d[0][j]=='.'&&d[1][j]=='.'; j + +);113             if(j==i)Continue; the             if(i==0&&j==n) Ans^=sg.sg0 (J-i); the             Else if(i==0|| J==n) ANS^=SG.SG1 (J-i); the             Else if(d[0][i-1]^d[0][J]) ANS^=SG.SG2 (J-i);117             ElseANS^=SG.SG3 (J-i);118I=J;119         } - 121         returnAns?Win:lose;122     }123};

TC SRM 637 DIV1 pathgame