[SDOI2011]染色 题解

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）

思路:

　　树剖之后，维护其两端的颜色、答案和标记即可。

代码：

#include<cstdio>
#include<iostream>
#define N 100001
using namespace std;
int n,m,cnt,dfn,vis[N],head[N],to[N<<],next[N<<],deep[N],size[N],top[N],id[N],v[N],pa[N],lazy[N<<],lc[N<<],rc[N<<],sum[N<<];

void ins(int x,int y)
{
     to[++cnt]=y,next[cnt]=head[x],head[x]=cnt;
}

void dfs1(int x)
{
    size[x]=vis[x]=;
    for (int i=head[x];i;i=next[i])
        if (!vis[to[i]])
        {
            deep[to[i]]=deep[x]+;
            pa[to[i]]=x;
            dfs1(to[i]);
            size[x]+=size[to[i]];
        }
}

void dfs2(int x,int chain)
{
    int k=,i;
    id[x]=++dfn;
    top[x]=chain;
    for (i=head[x];i;i=next[i])
        if (deep[to[i]]>deep[x] && size[to[i]]>size[k]) k=to[i];
    if (!k) return;
    dfs2(k,chain);
    for (i=head[x];i;i=next[i])
        if (deep[to[i]]>deep[x] && to[i]!=k) dfs2(to[i],to[i]);
}

void build(int l,int r,int cur)
{
    sum[cur]=,lazy[cur]=-;
    if(l==r)return;
    int mid=l+r>>;
    build(l,mid,cur<<),build(mid+,r,cur<<|);
}

void update(int k)
{
    lc[k]=lc[k<<],rc[k]=rc[k<<|];
    if (rc[k<<]^lc[k<<|]) sum[k]=sum[k<<]+sum[k<<|];
    else sum[k]=sum[k<<]+sum[k<<|]-;
}

void pushdown(int l,int r,int k)
{
    int tmp=lazy[k]; lazy[k]=-;
    if (tmp==- || l==r) return;
    sum[k<<]=sum[k<<|]=;
    lazy[k<<]=lazy[k<<|]=tmp;
    lc[k<<]=rc[k<<]=tmp;
    lc[k<<|]=rc[k<<|]=tmp;
}

void change(int L,int R,int l,int r,int cur,int val)
{
    if (l==L && r==R) { lc[cur]=rc[cur]=val; sum[cur]=; lazy[cur]=val; return; }
    int mid=L+R>>; pushdown(L,R,cur);
    if (r<=mid) change(L,mid,l,r,cur<<,val);
    else if (l>mid) change(mid+,R,l,r,cur<<|,val);
         else change(L,mid,l,mid,cur<<,val),change(mid+,R,mid+,r,cur<<|,val);
    update(cur);
}

int ask(int L,int R,int l,int r,int cur)
{
    if (l==L && r==R) return sum[cur];
    int mid=L+R>>; pushdown(L,R,cur);
    if (r<=mid) return ask(L,mid,l,r,cur<<);
    else if (l>mid) return ask(mid+,R,l,r,cur<<|);
         else
         {
              int tmp=;
              if (rc[cur<<]^lc[cur<<|]) tmp=;
              return ask(L,mid,l,mid,cur<<)+ask(mid+,R,mid+,r,cur<<|)-tmp;
         }
}

int getc(int l,int r,int cur,int x)
{
    if (l==r) return lc[cur];
    int mid=l+r>>; pushdown(l,r,cur);
    if (x<=mid) return getc(l,mid,cur<<,x);
    else return getc(mid+,r,cur<<|,x);
}

int solvesum(int x,int y)
{
    int sum=;
    for (;top[x]!=top[y];x=pa[top[x]])
    {
        if (deep[top[x]]<deep[top[y]]) swap(x,y);
        sum+=ask(,n,id[top[x]],id[x],);
        if (getc(,n,,id[top[x]])==getc(,n,,id[pa[top[x]]])) sum--;
    }
    if (deep[x]>deep[y]) swap(x,y);
    return sum+=ask(,n,id[x],id[y],);
}

void solvechange(int x,int y,int val)
{
    for (;top[x]!=top[y];x=pa[top[x]])
    {
        if (deep[top[x]]<deep[top[y]]) swap(x,y);
        change(,n,id[top[x]],id[x],,val);
    }
    if (deep[x]>deep[y]) swap(x,y);
    change(,n,id[x],id[y],,val);
}

int main()
{
    int i,a,b,c;
    scanf("%d%d",&n,&m);
    for(i=;i<=n;i++)scanf("%d",&v[i]);
    for(i=;i<n;i++) scanf("%d%d",&a,&b),ins(a,b),ins(b,a);
    dfs1(),dfs2(,),build(,n,);
    for(i=;i<=n;i++) change(,n,id[i],id[i],,v[i]);
    for(i=;i<=m;i++)
    {
        char ch[];
        scanf("%s",ch);
        if(ch[]=='Q')
        {
            scanf("%d%d",&a,&b);
            printf("%d\n",solvesum(a,b));
        }
        else
        {
            scanf("%d%d%d",&a,&b,&c);
            solvechange(a,b,c);
        }
    }
    return ;
}