Bzoj 2588 count on a tree chair tree + multiply LCA

Given a tree, each node has the right value. Ask if the K value of the two node paths is smaller

This question is very slow...

Tree chain partitioning + binary + tree-nested O (nlog ^ 4N) practices can be eliminated

No modification operation. Tree link splitting + binary + tree division O (nlog ^ 3n) is still dead.

I am angry. I learned how to make a persistent line segment tree (without looking at any code otz, how can I achieve 0.0), binary + Chairman tree, O (nlog ^ 2n ), it's still dead!

Finally, I found that I am Sb, and there is no need to divide it into two parts. Just pass all the four parameters, O (nlogn)

First, we maintain the weight line segment tree from this node to the root for each node, and then for each query (x, y ), the line segment tree in this path is tree [x] + tree [y]-tree [LCA (x, y)]-tree [Fa [LCA (x, y)]

Pass all the four nodes as parameters and use them as the normal weight line segment tree to perform a normal k-small query.

Time complexity O (nlogn)

The last carriage return cannot be output, otherwise it will be pe... No wonder the number of PES in this question accounts for 2/3 of AC...

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define M 100100using namespace std;struct Tree{Tree *ls,*rs;int num;Tree(Tree*_,Tree*__,int ___):ls(_),rs(__),num(___){}}*tree[M];Tree* Build_Tree(Tree *p,int x,int y,int z){int mid=x+y>>1;if(x==y)return new Tree(tree[0],tree[0],p->num+1);if(z<=mid)return new Tree(Build_Tree(p->ls,x,mid,z),p->rs,p->num+1);elsereturn new Tree(p->ls,Build_Tree(p->rs,mid+1,y,z),p->num+1);}int Get_Ans(Tree *p1,Tree *p2,Tree *p3,Tree *p4,int x,int y,int k){int mid=x+y>>1;if(x==y)return mid;int temp = p1->ls->num + p2->ls->num - p3->ls->num - p4->ls->num ;if(k<=temp) return Get_Ans( p1->ls , p2->ls , p3->ls , p4->ls , x , mid , k );else return Get_Ans( p1->rs , p2->rs , p3->rs , p4->rs , mid+1 , y , k-temp );}struct abcd{int to,next;}table[M<<1];int head[M],tot;int n,m,ans;int a[M],fa[M][20],dpt[M];pair<int,int>b[M];void Add(int x,int y){table[++tot].to=y;table[tot].next=head[x];head[x]=tot;}void DFS(int x){int i;dpt[x]=dpt[fa[x][0]]+1;tree[x]=Build_Tree(tree[fa[x][0]],1,n,a[x]);for(i=head[x];i;i=table[i].next){if(table[i].to==fa[x][0])continue;fa[table[i].to][0]=x;DFS(table[i].to);}}int LCA(int x,int y){int j;if(dpt[x]<dpt[y])swap(x,y);for(j=19;~j;j--)if(dpt[fa[x][j]]>=dpt[y])x=fa[x][j];if(x==y)return x;for(j=19;~j;j--)if(fa[x][j]!=fa[y][j])x=fa[x][j],y=fa[y][j];return fa[x][0];}int Query(int x,int y,int k){int lca=LCA(x,y);return Get_Ans( tree[x] , tree[y] , tree[lca] , tree[fa[lca][0]] , 1 , n , k );}int main(){//freopen("count.in","r",stdin);//freopen("count.out","w",stdout);int i,j,x,y,k;cin>>n>>m;for(i=1;i<=n;i++){scanf("%d",&b[i].first);b[i].second=i;}sort(b+1,b+n+1);for(i=1;i<=n;i++)a[b[i].second]=i;for(i=1;i<n;i++){scanf("%d%d",&x,&y);Add(x,y);Add(y,x);}tree[0]=new Tree(0x0,0x0,0);tree[0]->ls=tree[0]->rs=tree[0];DFS(1);for(j=1;j<=19;j++)for(i=1;i<=n;i++)fa[i][j]=fa[ fa[i][j-1] ][j-1];for(i=1;i<=m;i++){scanf("%d%d%d",&x,&y,&k);printf("%d",ans=b[Query(x^ans,y,k)].first);if(i!=m)puts("");}}

The tree link splitting + tree partitioning provided by tle...

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define M 100100using namespace std;inline int getc() {    static const int L = 1 << 15;    static char buf[L], *S = buf, *T = buf;    if (S == T) {        T = (S = buf) + fread(buf, 1, L, stdin);        if (S == T)            return EOF;    }    return *S++;}inline int getint() {    int c;    while(!isdigit(c = getc()) && c != '-');    bool sign = c == '-';    int tmp = sign ? 0 : c - '0';    while(isdigit(c = getc()))        tmp = (tmp << 1) + (tmp << 3) + c - '0';    return sign ? -tmp : tmp;}inline void output(int x) {    static int a[20];    if (x == 0)        putchar('0');    else {        int top = 0;        if (x < 0)            putchar('-'), x=-x;        while(x) {            a[++top] = x % 10;            x /= 10;        }        for(int i = top; i >= 1; --i)            putchar('0' + a[i]);    }}struct abcd{    int to,next;}table[M<<1];int head[M],tot;int n,m,ans,maxnum;int f[M],fa[M],son[M],dpt[M],siz[M],top[M],pos[M],a[M],b[M],c[M],s[20][M],cnt;inline void add(int x,int y){    table[++tot].to=y;    table[tot].next=head[x];    head[x]=tot;}void bfs(){    static int q[M],r=0,h=0;    int i,x;    q[++r]=1;    while(r!=h)    {        x=q[++h];        dpt[x]=dpt[fa[x]]+1;        siz[x]=1;        for(i=head[x];i;i=table[i].next)        {            if(table[i].to==fa[x])                continue;            fa[table[i].to]=x;            q[++r]=table[i].to;        }    }    for(i=n;i;i--)    {        x=q[i];        siz[fa[x]]+=siz[x];        if(siz[x]>siz[son[fa[x]]])            son[fa[x]]=x;    }    for(i=1;i<=n;i++)    {        x=q[i];        if(son[fa[x]]==x)            top[x]=top[fa[x]];        else        {            top[x]=x;            for(;x;x=son[x])                pos[x]=++cnt,a[pos[x]]=f[x];        }    }}void Build_Tree(int l,int r,int dpt){    if(l==r)        return ;    int i,mid=l+r>>1;    int l1=l,l2=mid+1;    int left=mid-l+1;    for(i=l;i<=r;i++)        left-=(a[i]<c[mid]);    for(i=l;i<=r;i++)    {        if(a[i]<c[mid]||a[i]==c[mid]&&left)            b[l1++]=a[i],s[dpt][i]=(i==l?1:s[dpt][i-1]+1),left-=(a[i]==c[mid]);        else            b[l2++]=a[i],s[dpt][i]=(i==l?0:s[dpt][i-1]);    }    memcpy( a+l , b+l , sizeof(a[0])*(r-l+1) );    Build_Tree(l,mid,dpt+1);    Build_Tree(mid+1,r,dpt+1);}int Get_Ans(int l,int r,int dpt,int x,int y,int val){    int mid=l+r>>1;    int l1=(x==l?0:s[dpt][x-1]),l2=s[dpt][y];    if(x>y)        return 0;    if(l==r)        return c[mid]<=val;    if(val<c[mid])        return Get_Ans(l,mid,dpt+1,l+l1,l+l2-1,val);    else        return l2-l1+Get_Ans(mid+1,r,dpt+1,(mid+1)+(x-l-l1),(mid+1)+(y-l+1-l2)-1,val);}bool Query(int x,int y,int val,int k){    int re=0,fx=top[x],fy=top[y];    while(fx!=fy)    {        if(dpt[fx]<dpt[fy])            swap(x,y),swap(fx,fy);        re+=Get_Ans(1,n,0,pos[fx],pos[x],val);        if(re>=k)            return true;        x=fa[fx];fx=top[x];    }    if(dpt[x]<dpt[y])        swap(x,y);    re+=Get_Ans(1,n,0,pos[y],pos[x],val);    if(re>=k)        return true;    return false;}inline int Divide(int x,int y,int k){    int l=0,r=maxnum;    while(l+1!=r)    {        int mid=l+r>>1;        if( Query(x,y,mid,k) )            r=mid;        else            l=mid;    }    if( Query(x,y,l,k) )        return l;    return r;}int main(){//freopen("count.in","r",stdin);//freopen("bf.out","w",stdout);    int i,x,y,k;    cin>>n>>m;    for(i=1;i<=n;i++)        f[i]=getint(),maxnum=max(maxnum,f[i]);    for(i=1;i<n;i++)        x=getint(),y=getint(),add(x,y),add(y,x);    bfs();    memcpy(c+1,a+1,n<<2);    sort(c+1,c+n+1);    Build_Tree(1,n,0);    for(i=1;i<=m;i++)    {        x=getint();y=getint();k=getint();        x^=ans;        ans=Divide(x,y,k);        output(ans);        if(i!=m)        puts("");    }    return 0;}

bzoj 2588 count on a tree chair tree + multiply LCA