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<<1],next[N<<1],deep[N],size[N],top[N],id[N],v[N],pa[N],lazy[N<<2],lc[N<<2],rc[N<<2],sum[N<<2];
void ins(int x,int y)
{
to[++cnt]=y,next[cnt]=head[x],head[x]=cnt;
}
void dfs1(int x)
{
size[x]=vis[x]=1;
for (int i=head[x];i;i=next[i])
if (!vis[to[i]])
{
deep[to[i]]=deep[x]+1;
pa[to[i]]=x;
dfs1(to[i]);
size[x]+=size[to[i]];
}
}
void dfs2(int x,int chain)
{
int k=0,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]=1,lazy[cur]=-1;
if(l==r)return;
int mid=l+r>>1;
build(l,mid,cur<<1),build(mid+1,r,cur<<1|1);
}
void update(int k)
{
lc[k]=lc[k<<1],rc[k]=rc[k<<1|1];
if (rc[k<<1]^lc[k<<1|1]) sum[k]=sum[k<<1]+sum[k<<1|1];
else sum[k]=sum[k<<1]+sum[k<<1|1]-1;
}
void pushdown(int l,int r,int k)
{
int tmp=lazy[k]; lazy[k]=-1;
if (tmp==-1 || l==r) return;
sum[k<<1]=sum[k<<1|1]=1;
lazy[k<<1]=lazy[k<<1|1]=tmp;
lc[k<<1]=rc[k<<1]=tmp;
lc[k<<1|1]=rc[k<<1|1]=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]=1; lazy[cur]=val; return; }
int mid=L+R>>1; pushdown(L,R,cur);
if (r<=mid) change(L,mid,l,r,cur<<1,val);
else if (l>mid) change(mid+1,R,l,r,cur<<1|1,val);
else change(L,mid,l,mid,cur<<1,val),change(mid+1,R,mid+1,r,cur<<1|1,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>>1; pushdown(L,R,cur);
if (r<=mid) return ask(L,mid,l,r,cur<<1);
else if (l>mid) return ask(mid+1,R,l,r,cur<<1|1);
else
{
int tmp=1;
if (rc[cur<<1]^lc[cur<<1|1]) tmp=0;
return ask(L,mid,l,mid,cur<<1)+ask(mid+1,R,mid+1,r,cur<<1|1)-tmp;
}
}
int getc(int l,int r,int cur,int x)
{
if (l==r) return lc[cur];
int mid=l+r>>1; pushdown(l,r,cur);
if (x<=mid) return getc(l,mid,cur<<1,x);
else return getc(mid+1,r,cur<<1|1,x);
}
int solvesum(int x,int y)
{
int sum=0;
for (;top[x]!=top[y];x=pa[top[x]])
{
if (deep[top[x]]<deep[top[y]]) swap(x,y);
sum+=ask(1,n,id[top[x]],id[x],1);
if (getc(1,n,1,id[top[x]])==getc(1,n,1,id[pa[top[x]]])) sum--;
}
if (deep[x]>deep[y]) swap(x,y);
return sum+=ask(1,n,id[x],id[y],1);
}
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(1,n,id[top[x]],id[x],1,val);
}
if (deep[x]>deep[y]) swap(x,y);
change(1,n,id[x],id[y],1,val);
}
int main()
{
int i,a,b,c;
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++)scanf("%d",&v[i]);
for(i=1;i<n;i++) scanf("%d%d",&a,&b),ins(a,b),ins(b,a);
dfs1(1),dfs2(1,1),build(1,n,1);
for(i=1;i<=n;i++) change(1,n,id[i],id[i],1,v[i]);
for(i=1;i<=m;i++)
{
char ch[10];
scanf("%s",ch);
if(ch[0]=='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 0;
}
鄂公网安备 42010202000505 号