SPOJ QTree6 [树链剖分]

啊啊啊！！！
maintain的时候忘记+1s+1，于是就。。。累死了，只剩下最后一个Qtree7.
这道题与上道题类似，关键在于路径上面判定同色，关键点在findpath上，至于线段树，我又一次惊叹于线段树在序列问题上近乎无敌的维护效率。中间的mid与mid+1的判定神来之笔。
看来我就是一只菜鸡。

#include<deque>#include<vector>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int N = 100100;#define mid ((l+r)>>1)struct data{    int next,to;}E[N<<2];struct node{    int l,r,ret;    node(int l=0,int r=0,int s=1):l(l),r(r),ret(s){}}tree[N<<2];bool G,H;int n,q,x,y,z;int head[N],sz[N],top[N],dfn[N],tot,cnt,fa[N],dft[N],sum[N],son[N];int ls[N<<2],rs[N<<2],root[N],val[N][2],col[N],bct;inline void addedge(int x,int y){    E[++tot].to=y,E[tot].next=head[x],head[x]=tot;    E[++tot].to=x,E[tot].next=head[y],head[y]=tot;}inline void dfsf(int u){    sz[u]=1;    for(register int v,i=head[u];i;i=E[i].next)        if(v=E[i].to,v!=fa[u]){            fa[v]=u,dfsf(v),sz[u]+=sz[v];            if(sz[son[u]]<sz[v])son[u]=v;        }}inline void dfss(int u,int f){    top[u]=f,dft[dfn[u]=++cnt]=u,sum[f]++;    if(son[u])dfss(son[u],f);    for(register int v,i=head[u];i;i=E[i].next)        if(v=E[i].to,v!=fa[u]&&v!=son[u])dfss(v,v);}void maintain(int rt,int x){    tree[rt].l=tree[rt].r=val[x][col[x]]+1;}inline node merge(node &x,node &y,bool b){    node rt=node(x.l,y.r);    if(b&&x.ret)rt.l=x.l+y.l;    if(b&&y.ret)rt.r=y.r+x.r;    rt.ret=b&&y.ret&&x.ret;    return rt;}inline void build(int rt,int l,int r){    if(l==r){        int u=dft[l];        for(register int v,i=head[u];i;i=E[i].next)        if(v=E[i].to,v!=fa[u]&&top[u]!=top[v]){                build(root[v]=++bct,dfn[v],dfn[v]+sum[v]-1),                val[u][col[v]]+=tree[root[v]].l;            }        maintain(rt,u);    }else{        build(ls[rt]=++bct,l,mid),build(rs[rt]=++bct,mid+1,r);        tree[rt]=merge(tree[ls[rt]],tree[rs[rt]],col[dft[mid]]==col[dft[mid+1]]);    }}inline bool judge(int rt,int l,int r,int L,int R,int ret=1){    if(l==r)return ret;    if(L<=mid)ret&=judge(ls[rt],l,mid,L,R);    if(mid<R)ret&=judge(rs[rt],mid+1,r,L,R);    if(L<=mid&&R>mid)ret&=col[dft[mid]]==col[dft[mid+1]];    return ret;}deque<int> Q;inline void findpath(int x,bool flag){    Q.clear();    while(x){        int f=top[x];        if(flag){            if(!judge(root[f],dfn[f],dfn[x]+sum[f]-1,dfn[f],dfn[x])||col[fa[f]]!=col[f]||!fa[f]){                Q.push_front(x),Q.push_front(f);break;            }        }else Q.push_front(x),Q.push_front(f);        x=fa[f];    }}inline void update(int rt,int l,int r,int x){    if(l==r){        int f=dft[l];        if(x+2<Q.size()){            int t=Q[x+1];            val[f][col[t]]-=tree[root[t]].l;            update(root[t],dfn[t],dfn[t]+sum[t]-1,x+2);            val[f][col[t]]+=tree[root[t]].l;        }else col[f]=1-col[f];        maintain(rt,f);    }else{        if(dfn[Q[x]]<=mid)update(ls[rt],l,mid,x);        else update(rs[rt],mid+1,r,x);        tree[rt]=merge(tree[ls[rt]],tree[rs[rt]],col[dft[mid]]==col[dft[mid+1]]);    }}inline int query(int rt,int l,int r,int x,bool &L,bool &R,int res=0){    if(l==r)return tree[rt].l;    if(dfn[Q[x]]<=mid){        res+=query(ls[rt],l,mid,x,L,R);        if(R&&col[dft[mid+1]]==col[dft[mid]])res+=tree[rs[rt]].l;        R=R&&tree[rs[rt]].ret&&col[dft[mid+1]]==col[dft[mid]];    }else{        res+=query(rs[rt],mid+1,r,x,L,R);        if(L&&col[dft[mid+1]]==col[dft[mid]])res+=tree[ls[rt]].r;        L=L&&tree[ls[rt]].ret&&col[dft[mid+1]]==col[dft[mid]];    }    return res;}inline void change(int u){    findpath(u,0);    update(1,1,sum[1],1);}inline void ask(int u){    findpath(u,1),u=Q[0];    printf("%d\n",query(root[u],dfn[u],dfn[u]+sum[u]-1,1,G=1,H=1));}inline void read(int &res){    static char ch;int flag=1;    while((ch=getchar())<'0'||ch>'9')if(ch=='-')flag=-1;res=ch-48;    while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-48;res*=flag;}int main(){    read(n);    for(register int i=1;i<n;i++)read(x),read(y),addedge(x,y);    read(q),dfsf(1),dfss(1,1);    build(root[1]=++bct,1,sum[1]);    for(register int i=1;i<=q;++i){        read(x),read(y);        if(x)change(y);        else ask(y);    }    return 0;}