[C++] BZOJ 1036: [ZJOI2008]树的统计Count 树链剖分

//qscqesze#include <cstdio>#include <cmath>#include <cstring>#include <ctime>#include <iostream>#include <algorithm>#include <set>#include <vector>#include <sstream>#include <queue>#include <typeinfo>#include <fstream>#include <map>#include <stack>typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define maxn 200001#define mod 10007#define eps 1e-9int Num;char CH[20];//const int inf=0x7fffffff;   //нчоч╢Сconst int inf=0x3f3f3f3f;/*inline void P(int x){    Num=0;if(!x){putchar('0');puts("");return;}    while(x>0)CH[++Num]=x%10,x/=10;    while(Num)putchar(CH[Num--]+48);    puts("");}*/inline ll read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}inline void P(int x){    Num=0;if(!x){putchar('0');puts("");return;}    while(x>0)CH[++Num]=x%10,x/=10;    while(Num)putchar(CH[Num--]+48);    puts("");}//**************************************************************************************#define N 1000000#define M 1000000int n,q,cnt,sz;int fa[N][15],v[N],deep[N],size[N],head[N];int pos[N],belong[N];bool vis[N];struct data{    int to,next;}e[M];struct seg{    int l,r,mx,sum;}t[100005];void insert(int u,int v){    e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt;    e[++cnt].to=u;e[cnt].next=head[v];head[v]=cnt;}void ini(){    scanf("%d",&n);    for(int i=1;i<n;i++)    {        int x,y;        scanf("%d%d",&x,&y);        insert(x,y);    }    for(int i=1;i<=n;i++)        scanf("%d",&v[i]);}void dfs1(int x){    size[x]=1;    vis[x]=1;    for(int i=1;i<=14;i++)    {        if(deep[x]<(1<<i))            break;        fa[x][i]=fa[fa[x][i-1]][i-1];    }    for(int i=head[x];i;i=e[i].next)    {        if(vis[e[i].to])            continue;        deep[e[i].to]=deep[x]+1;        fa[e[i].to][0]=x;        dfs1(e[i].to);        size[x]+=size[e[i].to];    }}//第一遍dfs求出树每个结点的深度deep[x]，其为根的子树大小size[x]//以及祖先的信息fa[x][i]表示x往上距离为2^i的祖先void dfs2(int x,int chain){    int k=0;    sz++;    pos[x]=sz;    belong[x]=chain;    for(int i=head[x];i;i=e[i].next)        if(deep[e[i].to]>deep[x]&&size[e[i].to]>size[k])            k=e[i].to;    if(k==0)        return;    dfs2(k,chain);    for(int i=head[x];i;i=e[i].next)        if(deep[e[i].to]>deep[x]&&k!=e[i].to)            dfs2(e[i].to,e[i].to);}int lca(int x,int y){    if(deep[x]<deep[y])        swap(x,y);    int t=deep[x]-deep[y];    for(int i=0;i<=14;i++)        if(t&(1<<i))            x=fa[x][i];    for(int i=14;i>=0;i--)    {        if(fa[x][i]!=fa[y][i])        {            x=fa[x][i];            y=fa[y][i];        }    }    if(x==y)        return x;    else        return fa[x][0];}void build(int k,int l,int r){    t[k].l=l,t[k].r=r;    if(l==r)        return ;    int mid=(l+r)>>1;    build(k<<1,l,mid);    build(k<<1|1,mid+1,r);}void change(int k,int x,int y){    int l=t[k].l,r=t[k].r,mid=(l+r)>>1;    if(l==r)    {        t[k].sum=t[k].mx=y;        return;    }    if(x<=mid)        change(k<<1,x,y);    else        change(k<<1|1,x,y);    t[k].sum=t[k<<1].sum+t[k<<1|1].sum;    t[k].mx=max(t[k<<1].mx,t[k<<1|1].mx);}int querysum(int k,int x,int y){    int l=t[k].l,r=t[k].r,mid=(l+r)>>1;    if(l==x&&y==r)        return t[k].sum;    if(y<=mid)return querysum(k<<1,x,y);    else if(x>mid)return querysum(k<<1|1,x,y);    else return querysum(k<<1,x,mid)+querysum(k<<1|1,mid+1,y);}int querymx(int k,int x,int y){    int l=t[k].l,r=t[k].r,mid=(l+r)>>1;    if(l==x&&y==r)return t[k].mx;    if(y<=mid)return querymx(k<<1,x,y);    else if(x>mid)return querymx(k<<1|1,x,y);    else return max(querymx(k<<1,x,mid),querymx(k<<1|1,mid+1,y));}int solvesum(int x,int f){    int sum=0;    while(belong[x]!=belong[f])    {        sum+=querysum(1,pos[belong[x]],pos[x]);        x=fa[belong[x]][0];    }    sum+=querysum(1,pos[f],pos[x]);    return sum;}int solvemx(int x,int f){    int mx=-inf;    while(belong[x]!=belong[f])    {        mx=max(mx,querymx(1,pos[belong[x]],pos[x]));        x=fa[belong[x]][0];    }    mx=max(mx,querymx(1,pos[f],pos[x]));    return mx;}void solve(){    build(1,1,n);    for(int i=1;i<=n;i++)        change(1,pos[i],v[i]);    scanf("%d",&q);    char ch[6];    for(int i=1;i<=q;i++)    {        int x,y;        scanf("%s%d%d",ch,&x,&y);        if(ch[0]=='C')        {            v[x]=y;            change(1,pos[x],y);        }        else        {            int t=lca(x,y);            if(ch[1]=='M')                printf("%d\n",max(solvemx(x,t),solvemx(y,t)));            else                printf("%d\n",solvesum(x,t)+solvesum(y,t)-v[t]);        }    }}int main(){    ini();    dfs1(1);    dfs2(1,1);    solve();}