【Luogu】U16325小奇的花园（树链剖分）

　　题目链接

　　学了学动态开点的树链剖分，其实跟动态开点的线段树差不多啦

　　查询的时候别ssbb地动态开点，如果没这个点果断返回0就行

　　只要注意花的种类能到intmax就行qwq！！！！

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<cstdlib>
#include<map>
#define mid ((l+r)>>1)
#define maxn 100010
#define check(x) if(x==0) x=++tot;
using namespace std;
inline long long read(){
    long long num=0,f=1;
    char ch=getchar();
    while(!isdigit(ch)){
        if(ch==‘-‘)    f=-1;
        ch=getchar();
    }
    while(isdigit(ch)){
        num=num*10+ch-‘0‘;
        ch=getchar();
    }
    return num*f;
}

int tot;
map<int,int> root;
int ls[maxn*100];
int rs[maxn*100];
int q[maxn];
int size[maxn];
int top[maxn];
int son[maxn];
int father[maxn];
int dfn[maxn];
int deep[maxn];
int back[maxn],cnt;
int n,m;

struct Edge{
    int next,to;
}edge[maxn*3];
int head[maxn],num;
inline void add(int from,int to){
    edge[++num]=(Edge){head[from],to};
    head[from]=num;
}

void find(int x,int fa){
    size[x]=1;    deep[x]=deep[fa]+1;
    for(int i=head[x];i;i=edge[i].next){
        int to=edge[i].to;
        if(to==fa)    continue;
        father[to]=x;
        find(to,x);
        size[x]+=size[to];
        if(son[x]==0||size[son[x]]<size[to])    son[x]=to;
    }
}

void unionn(int x,int Top){
    top[x]=Top;    dfn[x]=++cnt;    back[cnt]=x;
    if(son[x]==0)    return;
    unionn(son[x],Top);
    for(int i=head[x];i;i=edge[i].next){
        int to=edge[i].to;
        if(to==father[x]||to==son[x])    continue;
        unionn(to,to);
    }
    return;
}

inline void pushup(int rt){    tree[rt]=tree[ls[rt]]+tree[rs[rt]];    }

void update(int o,int num,int l,int r,int &rt){
    check(rt);
    if(l==r){
        tree[rt]+=num;
        return;
    }
    if(o<=mid)    update(o,num,l,mid,ls[rt]);
    else        update(o,num,mid+1,r,rs[rt]);
    pushup(rt);
}

int query(int from,int to,int l,int r,int &rt){
    if(rt==0)    return 0;
    if(from<=l&&to>=r)    return tree[rt];
    int ans=0;
    if(from<=mid)    ans+=query(from,to,l,mid,ls[rt]);
    if(to>mid)        ans+=query(from,to,mid+1,r,rs[rt]);
    return ans;
}

inline void addcol(int pos,int val){
    update(dfn[pos],-1,1,n,root[q[pos]]);
    update(dfn[pos],1,1,n,root[val]);
    q[pos]=val;
}

int ask(int from,int to,int val){
    int ans=0;
    while(top[from]!=top[to]){
        if(deep[top[from]]<deep[top[to]])    swap(from,to);
        ans+=query(dfn[top[from]],dfn[from],1,n,root[val]);
        from=father[top[from]];
    }
    if(deep[from]>deep[to])    swap(from,to);
    ans+=query(dfn[from],dfn[to],1,n,root[val]);
    return ans;
}

int main(){
    n=read(),m=read();
    for(int i=1;i<=n;++i)    q[i]=read();
    for(int i=1;i<n;++i){
        int from=read(),to=read();
        add(from,to);
        add(to,from);
    }
    find(1,1);
    unionn(1,1);
    for(int i=1;i<=n;++i)    update(dfn[i],1,1,n,root[q[i]]);
    int last=0;
    for(int i=1;i<=m;++i){
        char c[10];int x,y;
        scanf("%s",c);
        if(c[0]==‘C‘){
            x=read()^last,y=read()^last;
            addcol(x,y);
        }
        else{
            x=read()^last,y=read()^last;
            int z=read()^last;
            last=ask(x,y,z);
            printf("%d\n",last);
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/cellular-automaton/p/8352260.html