树链剖分套线段树
题目链接：bzoj 2243
题意：写 B z o j 的题意就是简单——题意见上……
题解：查询合并时注意记录上一次的颜色，交换 X、Y 时注意同样需要交换颜色哦。
我思故我在：树链剖分的水题，难得数据结构题没有花时间去调，唯一一处错误还是 lazy 标记的数组开小了。
bzoj 2243 染色
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//by yyh
//to sn
using namespace std;
int n,m;
int head=0;
int sn,yyh;
int last1=-1,last2=-1;
int now1,now2;
int w[100100];
vector<int> ve[100100];
int deep[100100],size[100100];
int fa[100100],top[100100],son[100100];
int where[100100],here[100100];
int lazy[2600000];
struct table{
int all,lc,rc;
}tree[2600000];
void dfs1(int x)
{
int sum=0;
int maxa=-1;
for(int i=0;i<ve[x].size();i++)
{
int y=ve[x][i];
if(y!=fa[x])
{
deep[y]=deep[x]+1;
fa[y]=x;
dfs1(y);
sum+=size[y];
if(size[y]>maxa)
maxa=size[y],son[x]=y;
}
}
size[x]=sum+1;
}
void dfs2(int x)
{
here[++head]=x;
where[x]=head;
if(son[x]==0) return ;
top[son[x]]=top[x];
dfs2(son[x]);
for(int i=0;i<ve[x].size();i++)
{
int y=ve[x][i];
if(y!=fa[x]&&y!=son[x])
{
top[y]=y;
dfs2(y);
}
}
}
void dfs(int x,int l,int r)
{
if(l==r)
tree[x].all=1,tree[x].lc=tree[x].rc=w[here[l]];
else
{
int mid=(l+r)/2;
dfs(x*2,l,mid);
dfs(x*2+1,mid+1,r);
tree[x].all=tree[x*2].all+tree[x*2+1].all;
if(tree[x*2].rc==tree[x*2+1].lc) tree[x].all--;
tree[x].lc=tree[x*2].lc,tree[x].rc=tree[x*2+1].rc;
}
}
int ask(int x,int l,int r,int f,int t)
{
if(lazy[x]!=-1)
{
tree[x*2].all=1,tree[x*2].lc=tree[x*2].rc=lazy[x],lazy[x*2]=lazy[x];
tree[x*2+1].all=1,tree[x*2+1].lc=tree[x*2+1].rc=lazy[x],lazy[x*2+1]=lazy[x];
lazy[x]=-1;
}
if(l==f&&r==t)
{
if(l==sn) now1=tree[x].lc;
if(r==yyh) now2=tree[x].rc;
return tree[x].all;
}
else
{
int mid=(l+r)/2;
if(t<=mid)
{
return ask(x*2,l,mid,f,t);
}
else if(f>=mid+1)
{
return ask(x*2+1,mid+1,r,f,t);
}
else
{
return ask(x*2,l,mid,f,mid)+ask(x*2+1,mid+1,r,mid+1,t)-(tree[x*2].rc==tree[x*2+1].lc);
}
}
}
void add(int x,int l,int r,int f,int t,int g)
{
if(lazy[x]!=-1)
{
tree[x*2].all=1,tree[x*2].lc=tree[x*2].rc=lazy[x],lazy[x*2]=lazy[x];
tree[x*2+1].all=1,tree[x*2+1].lc=tree[x*2+1].rc=lazy[x],lazy[x*2+1]=lazy[x];
lazy[x]=-1;
}
if(l==f&&r==t)
tree[x].all=1,tree[x].lc=tree[x].rc=g,lazy[x]=g;
else
{
int mid=(l+r)/2;
if(t<=mid)
{
add(x*2,l,mid,f,t,g);
}
else if(f>=mid+1)
{
add(x*2+1,mid+1,r,f,t,g);
}
else
{
add(x*2,l,mid,f,mid,g);
add(x*2+1,mid+1,r,mid+1,t,g);
}
tree[x].all=tree[x*2].all+tree[x*2+1].all;
if(tree[x*2].rc==tree[x*2+1].lc) tree[x].all--;
tree[x].lc=tree[x*2].lc,tree[x].rc=tree[x*2+1].rc;
}
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",&w[i]);
for(int i=1;i<n;i++)
{
int x,y;
scanf("%d%d",&x,&y);
ve[x].push_back(y);
ve[y].push_back(x);
}
deep[1]=1,fa[1]=1;
dfs1(1);
top[1]=1;
dfs2(1);
dfs(1,1,head);
for(int i=1;i<=2000000;i++)
lazy[i]=-1;
while(m--)
{
char a[3];
scanf("%s",a);
if(a[0]=='Q')
{
int x,y;
scanf("%d%d",&x,&y);
int ans=0;
last1=last2=-1;
while(top[x]!=top[y])
{
if(deep[top[x]]<deep[top[y]]) swap(x,y),swap(last1,last2);
sn=where[top[x]],yyh=where[x];
ans+=ask(1,1,head,where[top[x]],where[x]);
if(last1==now2) ans--;
last1=now1;
x=fa[top[x]];
}
if(deep[x]>deep[y]) swap(x,y),swap(last1,last2);
sn=where[x],yyh=where[y];
ans+=ask(1,1,head,where[x],where[y]);
if(now1==last1) ans--;
if(now2==last2) ans--;
printf("%d\n",ans);
}
else
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
while(top[x]!=top[y])
{
if(deep[top[x]]<deep[top[y]]) swap(x,y);
add(1,1,head,where[top[x]],where[x],z);
x=fa[top[x]];
}
if(deep[x]>deep[y]) swap(x,y);
add(1,1,head,where[x],where[y],z);
}
}
return 0;
}