BZOJ 1453 Wc2005 Dface双面棋盘 Link-Cut-Tree

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题目大意：给定一张网格图，每个点有黑色和白色，同色相邻的点之间连边，每次反转一个点的颜色并输出黑色和白色的联通块个数
《论科技不发达的危害》
这显然是个动态图问题，由于不强制在线，我们用Link-Cut-Tree维护边删除时间的最大生成树就可以了
时间复杂度O((n2+m)logn)
常数巨大
#include <queue>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define M 40400#define P(x,y) ((x)*n-n+(y))using namespace std;const int dx[]={0,0,1,-1};const int dy[]={1,-1,0,0};int n,m,ans[2];int a[220][220];queue<int> ed[M<<1];pair<int,int> edges[M<<1];pair<int,int> operations[10100];namespace Link_Cut_Tree{    bool Compare(int x,int y)    {        if(!x) return false;        if(!y) return true;        return ed[x].front()<ed[y].front();    }    struct abcd{        abcd *ls,*rs,*fa;        int e,min_e;        bool rev_mark;        abcd(int _);        void Push_Up();        void Push_Down();        void Reverse();    }*null=new abcd(0),*tree[M],*edge[M<<1];    abcd :: abcd(int _)    {        ls=rs=fa=null;        e=min_e=_;        rev_mark=false;    }    void abcd :: Push_Up()    {        min_e=min(min(ls->min_e,rs->min_e,Compare),e,Compare);    }    void abcd :: Push_Down()    {        if(fa->ls==this||fa->rs==this)            fa->Push_Down();        if(rev_mark)        {            ls->Reverse();            rs->Reverse();            rev_mark=false;        }    }    void abcd :: Reverse()    {        swap(ls,rs);        rev_mark^=1;    }    void Zig(abcd *x)    {        abcd *y=x->fa;        y->ls=x->rs;        x->rs->fa=y;        x->rs=y;        x->fa=y->fa;        if(y==y->fa->ls)            y->fa->ls=x;        else if(y==y->fa->rs)            y->fa->rs=x;        y->fa=x;        y->Push_Up();    }    void Zag(abcd *x)    {        abcd *y=x->fa;        y->rs=x->ls;        x->ls->fa=y;        x->ls=y;        x->fa=y->fa;        if(y==y->fa->ls)            y->fa->ls=x;        else if(y==y->fa->rs)            y->fa->rs=x;        y->fa=x;        y->Push_Up();    }    void Splay(abcd *x)    {        x->Push_Down();        while(x->fa->ls==x||x->fa->rs==x)        {            abcd *y=x->fa,*z=y->fa;            if(x==y->ls)            {                if(y==z->ls)                    Zig(y);                Zig(x);            }            else            {                if(y==z->rs)                    Zag(y);                Zag(x);            }        }        x->Push_Up();    }    void Access(abcd *x)    {        abcd *y=null;        while(x!=null)        {            Splay(x);            x->rs=y;            x->Push_Up();            y=x;x=x->fa;        }    }    void Move_To_Root(abcd *x)    {        Access(x);        Splay(x);        x->Reverse();    }    abcd* Find_Root(abcd *x)    {        while(x->fa!=null)            x=x->fa;        return x;    }    bool Connect(abcd *x,abcd *y)    {        Access(x);        Splay(x);        bool re=Find_Root(y)==x;        Access(y);        return re;    }    void Link(abcd *x,abcd *y)    {        Move_To_Root(x);        x->fa=y;    }    bool Cut(abcd *x,abcd *y)    {        Move_To_Root(x);        Access(y);        Splay(y);        if( y->ls!=x || x->rs!=null )            return false;        y->ls=null;        x->fa=null;        y->Push_Up();        return true;    }}void Cut(int e,bool col){    using namespace Link_Cut_Tree;    int x=edges[e].first;    int y=edges[e].second;    if( Cut(edge[e],tree[x]) && Cut(edge[e],tree[y]) )        ++ans[col];    ed[e].pop();}void Link(int e,bool col){    using namespace Link_Cut_Tree;    int x=edges[e].first;    int y=edges[e].second;    if(!Connect(tree[x],tree[y]) )    {        Link(edge[e],tree[x]);        Link(edge[e],tree[y]);        --ans[col];        return ;    }    Move_To_Root(tree[x]);    Access(tree[y]);    Splay(tree[y]);    int _e=tree[y]->min_e;    if( Compare(_e,e) )    {        int _x=edges[_e].first;        int _y=edges[_e].second;        Cut(edge[_e],tree[_x]);        Cut(edge[_e],tree[_y]);        Link(edge[e],tree[x]);        Link(edge[e],tree[y]);    }}int main(){    using namespace Link_Cut_Tree;    int i,j,k,x,y;    cin>>n;    for(i=1;i<=n;i++)        for(j=1;j<=n;j++)            scanf("%d",&a[i][j]);    //读入    for(i=1;i<=n;i++)        for(j=1;j<n;j++)            edges[P(i,j)]=pair<int,int>(P(i,j),P(i,j+1));    for(i=1;i<n;i++)        for(j=1;j<=n;j++)            edges[P(i,j)+n*n]=pair<int,int>(P(i,j),P(i+1,j));    //设定每条边对应的两端点    cin>>m;    for(i=1;i<=m;i++)    {        scanf("%d%d",&x,&y);        operations[i].first=x;        operations[i].second=y;        for(k=0;k<4;k++)        {            int xx=x+dx[k];            int yy=y+dy[k];            if(!xx||!yy||xx==n+1||yy==n+1)                continue;            if(a[x][y]==a[xx][yy])                ed[min(P(x,y),P(xx,yy))+(k>>1)*n*n].push(i);        }        a[x][y]^=1;    }    //读入数据&&记录每条边的终止时刻    for(i=1;i<=n;i++)        for(j=1;j<=n;j++)            for(k=0;k<4;k+=2)            {                int xx=i+dx[k];                int yy=j+dy[k];                if(!xx||!yy||xx==n+1||yy==n+1)                    continue;                if(a[i][j]==a[xx][yy])                    ed[P(i,j)+(k>>1)*n*n].push(m+1);            }    //记录最后没有被删除的边的终止时刻    for(i=1;i<=n*n;i++)        tree[i]=new abcd(0);    for(i=1;i<=n;i++)        for(j=1;j<n;j++)            edge[P(i,j)]=new abcd(P(i,j));    for(i=1;i<n;i++)        for(j=1;j<=n;j++)            edge[P(i,j)+n*n]=new abcd(P(i,j)+n*n);    //申请内存    for(i=m;i;i--)    {        x=operations[i].first;        y=operations[i].second;        a[x][y]^=1;    }    //还原a数组    for(i=1;i<=n;i++)        for(j=1;j<=n;j++)            ans[a[i][j]]++;    for(i=1;i<=n;i++)        for(j=1;j<=n;j++)            for(k=0;k<4;k+=2)            {                int xx=i+dx[k];                int yy=j+dy[k];                if(!xx||!yy||xx==n+1||yy==n+1)                    continue;                if(a[i][j]==a[xx][yy])                    Link(P(i,j)+(k>>1)*n*n,a[i][j]);            }    //统计初始答案&&连接初始的边    for(i=1;i<=m;i++)    {        x=operations[i].first;        y=operations[i].second;        for(k=0;k<4;k++)        {            int xx=x+dx[k];            int yy=y+dy[k];            if(!xx||!yy||xx==n+1||yy==n+1)                continue;            if(a[x][y]==a[xx][yy])                Cut(min(P(x,y),P(xx,yy))+(k>>1)*n*n,a[x][y]);        }        ans[a[x][y]]--;        a[x][y]^=1;        ans[a[x][y]]++;        for(k=0;k<4;k++)        {            int xx=x+dx[k];            int yy=y+dy[k];            if(!xx||!yy||xx==n+1||yy==n+1)                continue;            if(a[x][y]==a[xx][yy])                Link(min(P(x,y),P(xx,yy))+(k>>1)*n*n,a[x][y]);        }        printf("%d %d\n",ans[1],ans[0]);    }    return 0;}
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