【bzoj2243】【树链剖分】【线段树】SDOI2011染色

• 题目大意：
  给定一棵有N 个节点的无根树和 M 个操作，操作有2类：
  
  1. 将节点A 到节点B路径上所有点都染成颜色 C
  2. 询问节点A到节点B路径上的颜色段数量（连续相同颜色被认为是同一段），如”112221”由3段组成:”11”.”222”和”1”。
     请你写一个程序依次完成这 M 个操作。
• 数据范围：
  

• 题解：
  对树上的路径进行操作， 区间覆盖，区间查询，一瞅就可以树剖，剖完之后用线段树维护区间某种颜色出现的个数col，区间左端点颜色lcol，区间右端点颜色rcol
• 合并时满足：
  
  p−>col=p−>lch−>col+p−>rch−>col−(p−>lch−>rcol==p−>rch−>lcol)
• 注意延重链向上爬的时候也要维护lcol和rcol进行统计答案
  直接上代码

#include <iostream>#include <cstring>#include <cstdio>#include <vector>#define MAXN 100001using namespace std;vector<int> e[MAXN];int n,m;int dep[MAXN],size[MAXN],hson[MAXN],fa[MAXN];void dfs1(int u,int f,int d){    size[u]=1,fa[u]=f,dep[u]=d;    for(vector<int>::iterator it=e[u].begin();it!=e[u].end();it++){        if(*it==f) continue;        dfs1(*it,u,d+1);        size[u]+=size[*it];        if(size[*it]>size[hson[u]]) hson[u]=*it;    }}int cnt[MAXN],top[MAXN],T,bel[MAXN];void dfs2(int u,int tp){    bel[T]=u,cnt[u]=T++,top[u]=tp;    if(hson[u]) dfs2(hson[u],tp);    for(vector<int>::iterator it=e[u].begin();it!=e[u].end();it++){        if(*it==hson[u]||*it==fa[u]) continue;        dfs2(*it,*it);    }}struct node{    int lnum,rnum,kind,l,r,mid,lzy;    node *lch,*rch;    node(){}    node(int a,int b,int c):l(a),r(b),mid(c){lnum=rnum=kind=lzy=0;lch=rch=NULL;}}*root;int C[MAXN];void push_down(node *p){    p->lch->lzy=p->rch->lzy=p->lzy;    p->lch->lnum=p->lch->rnum=p->lzy;    p->lch->kind=1;    p->rch->lnum=p->rch->rnum=p->lzy;    p->rch->kind=1;    p->lzy=0;}void push_up(node *p){    p->kind=p->lch->kind+p->rch->kind;    if(p->lch->rnum==p->rch->lnum)        p->kind--;    p->lnum=p->lch->lnum;    p->rnum=p->rch->rnum;}node *build(int left,int right){    node *p=new node(left,right,(left+right)>>1);    if(right-left==1){        p->kind=1;        p->lnum=p->rnum=C[bel[left]];        return p;    }    p->lch=build(left,p->mid);    p->rch=build(p->mid,right);    push_up(p);    return p;}void change(node *p,int left,int right,int col){    if(p->l==left&&p->r==right){        p->kind=1,p->lnum=p->rnum=col;        p->lzy=col;        return;    }    if(p->lzy)        push_down(p);    if(right<=p->mid)        change(p->lch,left,right,col);    else if(left>=p->mid)        change(p->rch,left,right,col);    else{        change(p->lch,left,p->mid,col);        change(p->rch,p->mid,right,col);    }    push_up(p);}struct Q{    int sum,lcol,rcol;    Q(int a,int b,int c):sum(a),lcol(b),rcol(c){}    Q(){}};Q query(node *p,int left,int right){    if(p->l==left&&p->r==right)        return Q(p->kind,p->lnum,p->rnum);    if(p->lzy)        push_down(p);    if(right<=p->mid)        return query(p->lch,left,right);    else if(left>=p->mid)        return query(p->rch,left,right);    else{        Q a=query(p->lch,left,p->mid);        Q b=query(p->rch,p->mid,right);        if(a.rcol==b.lcol)            return Q(a.sum+b.sum-1,a.lcol,b.rcol);        else            return Q(a.sum+b.sum,a.lcol,b.rcol);    }}void changeans(int u,int v,int c){    while(top[u]!=top[v]){        if(dep[top[u]]<dep[top[v]]) swap(u,v);        change(root,cnt[top[u]],cnt[u]+1,c);        u=fa[top[u]];    }    if(cnt[u]>cnt[v]) swap(u,v);    change(root,cnt[u],cnt[v]+1,c);}int getans(int s,int t){    pair<int,int> u,v;    u=make_pair(s,0);    v=make_pair(t,1);    int lc[2]={-1};    int ans[2]={0};    while(top[u.first]!=top[v.first]){        if(dep[top[u.first]]<dep[top[v.first]]) swap(u,v);        Q temp=query(root,cnt[top[u.first]],cnt[u.first]+1);        ans[u.second]=ans[u.second]+temp.sum-(lc[u.second]==temp.rcol);        lc[u.second]=temp.lcol;        u.first=fa[top[u.first]];    }    if(cnt[u.first]>cnt[v.first]) swap(u,v);    Q te=query(root,cnt[u.first],cnt[v.first]+1);    int anssum=ans[0]+ans[1]+te.sum-(lc[u.second]==te.lcol)-(lc[v.second]==te.rcol);    return anssum;}int main(){    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++)        scanf("%d",&C[i]);    for(int i=1;i<n;i++){        int a,b;        scanf("%d%d",&a,&b);        e[a].push_back(b);        e[b].push_back(a);    }    dfs1(1,-1,0);    dfs2(1,1);    root=build(0,T);    for(int i=1;i<=m;i++){        char a[2];        scanf("%s",a);        if(a[0]=='C'){            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            changeans(a,b,c);        }        else if(a[0]=='Q'){            int a,b;            scanf("%d%d",&a,&b);            printf("%d\n",getans(a,b));        }    }    return 0;}