BZOJ1453: [Wc]dface double-sided chessboard

Descriptioninputoutputsample Input
Sample Output
HINT

The line tree sets and the check set should be a better writing practice, the time complexity of O (N^3+m*nlogn).

#include <cstdio> #include <cctype> #include <queue> #include <cstring> #include <algorithm > #define REP (i,s,t) for (int. i=s;i<=t;i++) #define DWN (I,S,T) for (int. i=s;i>=t;i--) #define REN for (int i=first[x    ];i;i=next[i]) using namespace Std;inline int read () {int X=0,f=1;char c=getchar (); for (;!    IsDigit (c); C=getchar ()) if (c== '-') f=-1;    for (; IsDigit (c); C=getchar ()) x=x*10+c-' 0 '; return x*f;}    const int Maxn=210;int n,a[maxn][maxn];struct Node {int res[2],pa[maxn*2];    int findset (int x) {return pa[x]==x?x:findset (pa[x]);}        void init (int x) {res[0]=res[1]=0;        Rep (i,1,n) pa[i]=i; Rep (i,1,n) {res[!            a[x][i]]++; if (I>1&&a[x][i-1]==a[x][i]) res[!        A[x][i]]--, pa[findset (i-1)]=findset (i);    } Rep (i,1,n) pa[i+n]=pa[i]; }}t[maxn<<2];int pa[maxn*4],tmp[maxn*4];int findset (int x) {return x==pa[x]?x:pa[x]=findset (pa[x]);}    void Maintain (int o,int mid) {int lc=o<<1,rc=lc|1; T[o].rEs[0]=t[lc].res[0]+t[rc].res[0];    T[O].RES[1]=T[LC].RES[1]+T[RC].RES[1];    Rep (i,1,n*2) pa[i]=t[lc].pa[i];    Rep (i,1,n*2) pa[i+2*n]=t[rc].pa[i]+2*n;        Rep (i,1,n) if (A[mid][i]==a[mid+1][i]) {int X=findset (i+n), Y=findset (i+2*n); if (x!=y) {t[o].res[!            a[mid][i]]--;        Pa[x]=y;        }} Rep (i,1,n*4) {if (i<=n) tmp[findset (i)]=i;    if (i>3*n) tmp[findset (i)]=i-2*n;    } Rep (I,1,n) t[o].pa[i]=tmp[pa[i]]; Rep (I,1,n) t[o].pa[i+n]=tmp[pa[i+3*n]];}    void build (int o,int l,int R) {if (l==r) T[o].init (L);        else {int mid=l+r>>1,lc=o<<1,rc=lc|1;        Build (Lc,l,mid); build (Rc,mid+1,r);    Maintain (O,MID);    }}void Update (int o,int l,int R,int p) {if (l==r) T[o].init (L);        else {int mid=l+r>>1,lc=o<<1,rc=lc|1;        if (p<=mid) update (LC,L,MID,P);        else update (RC,MID+1,R,P);    Maintain (O,MID);    }}int Main () {n=read ();   Rep (I,1,n) Rep (j,1,n) A[i][j]=read (); Build (1,1,n); Dwn (I,read (), 1) {int x=read (), Y=read ();        A[x][y]^=1;        Update (1,1,N,X);    printf ("%d%d\n", t[1].res[0],t[1].res[1]); } return 0;}

　　

BZOJ1453: [Wc]dface double-sided chessboard