BZOJ2243
Description
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。
Input
第一行包含2个整数n和m，分别表示节点数和操作数；
第二行包含n个正整数表示n个节点的初始颜色
下面行每行包含两个整数x和y，表示x和y之间有一条无向边。
下面行每行描述一个操作：
“C a b c”表示这是一个染色操作，把节点a到节点b路径上所有点（包括a和b）都染成颜色c；
“Q a b”表示这是一个询问操作，询问节点a到节点b（包括a和b）路径上的颜色段数量。
Output
对于每个询问操作，输出一行答案。
Sample Input
6 5
2 2 1 2 1 1
1 2
1 3
2 4
2 5
2 6
Q 3 5
C 2 1 1
Q 3 5
C 5 1 2
Q 3 5
Sample Output
3
1
2
HINT
数N<=105，操作数M<=105，所有的颜色C为整数且在[0, 10^9]之间。
没什么好说的 但是为什么我的树链剖分长度总是比别人长
看来该压行的时候还是要压行
/*
if you can't see the repay
Why not just work step by step
rubbish is relaxed
to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
//#pragma comment(linker, "/STACK:10240000,10240000")
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
typedef long long ll;
typedef unsigned long long ull;
const ull hash1 = 201326611;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod = (int)1e9+7;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
//ull ha[MAX_N],pp[MAX_N];
inline int read()
{
int date = 0,m = 1; char ch = 0;
while(ch!='-'&&(ch<'0'|ch>'9'))ch = getchar();
if(ch=='-'){m = -1; ch = getchar();}
while(ch>='0' && ch<='9')
{
date = date*10+ch-'0';
ch = getchar();
}return date*m;
}
/*namespace sgt
{
#define mid ((l+r)>>1)
#undef mid
}*/
/*int root[MAX_N],cnt,sz;
namespace hjt
{
#define mid ((l+r)>>1)
struct node{int l,r,maxx;}T[MAX_N*40];
#undef mid
}*/
const int MAX_N = 100025;
int n,cnt,s[MAX_N<<2],rnum[MAX_N<<2],lnum[MAX_N<<2],col[MAX_N<<2],p[MAX_N],eid ,fa[MAX_N],sz[MAX_N],depth[MAX_N],son[MAX_N],Rank[MAX_N],top[MAX_N],id[MAX_N],arr[MAX_N];
void init()
{
memset(p,-1,sizeof(p));
eid = 0;
}
void up(int rt)
{
lnum[rt] = lnum[rt<<1];
rnum[rt] = rnum[rt<<1|1];
if(lnum[rt<<1|1]==rnum[rt<<1]) s[rt] = s[rt<<1]+s[rt<<1|1]-1;
else s[rt] = s[rt<<1] +s[rt<<1|1];
}
void build(int rt,int l,int r)
{
col[rt] = lnum[rt] = rnum[rt] = -1;s[rt] = 0;
if(l==r)
{
lnum[rt] = rnum[rt] = arr[Rank[l]];
s[rt] = 1;
return ;
}
int mid = (l+r)>>1;
build(rt<<1,l,mid);
build(rt<<1|1,mid+1,r);
up(rt);
}
void down(int rt,int l,int r)
{
if(col[rt]!=-1)
{
col[rt<<1] = col[rt<<1|1] = col[rt];
s[rt<<1] = s[rt<<1|1] = 1;
lnum[rt<<1] = lnum[rt<<1|1] = rnum[rt<<1] = rnum[rt<<1|1] = col[rt];
col[rt] = -1;
}
}
void update(int rt,int l,int r,int x,int y,int v)
{
if(x<=l&&r<=y)
{
lnum[rt] = rnum[rt] = col[rt] = v;
s[rt] = 1;
return ;
}
down(rt,l,r);
int mid = (l+r)>>1;
if(x>mid) update(rt<<1|1,mid+1,r,x,y,v);
else if(y<=mid) update(rt<<1,l,mid,x,y,v);
else update(rt<<1,l,mid,x,y,v),update(rt<<1|1,mid+1,r,x,y,v);
up(rt);
}
int query(int rt,int l,int r,int x,int y)
{
if(x<=l&&r<=y)
{
return s[rt];
}
int mid = (l+r)>>1;
down(rt,l,r);
if(x>mid) return query(rt<<1|1,mid+1,r,x,y);
else if(y<=mid) return query(rt<<1,l,mid,x,y);
else
{
if(lnum[rt<<1|1]==rnum[rt<<1]) return (query(rt<<1,l,mid,x,y)+query(rt<<1|1,mid+1,r,x,y) - 1);
else return (query(rt<<1,l,mid,x,y)+query(rt<<1|1,mid+1,r,x,y));
}
}
struct edge
{
int v,next;
}e[MAX_N<<1];
void add(int u,int v)
{
e[eid].v =v;
e[eid].next = p[u];
p[u] = eid++;
}
void dfs1(int x,int fath)
{
sz[x] = 1;
depth[x] = depth[fath]+1;
fa[x] = fath;
for(int i = p[x];i+1;i=e[i].next)
{
int v = e[i].v;
if(v==fath) continue;
dfs1(v,x);
sz[x]+=sz[v];
if(sz[v]>sz[son[x]]) son[x] = v;
}
}
void dfs2(int u,int t)
{
top[u] = t;
id[u] = ++cnt;
Rank[cnt] = u;
if(!son[u])
return;
dfs2(son[u],t);
for(int i = p[u];i+1;i=e[i].next)
{
int v = e[i].v;
if(v!=son[u]&&v!=fa[u])
dfs2(v,v);
}
}
void updates(int x,int y,int c)
{
while(top[x]!=top[y])
{
if(depth[top[x]]<depth[top[y]])
swap(x,y);
update(1,1,n,id[top[x]],id[x],c);
x = fa[top[x]];
}
if(id[x]>id[y])
swap(x,y);
update(1,1,n,id[x],id[y],c);
}
int getl(int rt,int l,int r,int x)
{
if(l==r)
{
return lnum[rt];
}
int mid = (l+r)>>1;
down(rt,l,r);
if(x<=mid) return getl(rt<<1,l,mid,x);
else return getl(rt<<1|1,mid+1,r,x);
}
inline int lca(int x,int y){
int f1=top[x],f2=top[y];
while(f1!=f2) {
if(depth[f1]<depth[f2]) swap(f1,f2),swap(x,y);
x=fa[f1]; f1=top[x];
}
if(depth[x]<depth[y]) swap(x,y);
return y;
}
int sum(int x,int y)
{
int ans = 0,last = -1,now = -1;
while(top[x]!=top[y])
{
now = getl(1,1,n,id[x]);
ans += query(1,1,n,id[top[x]],id[x]);
if(now==last) ans--;
last = getl(1,1,n,id[top[x]]);
x = fa[top[x]];
}
ans+=query(1,1,n,id[y],id[x]);
now = getl(1,1,n,id[x]);
if(now==last) ans--;
return ans;
}
int main()
{
//ios::sync_with_stdio(false);
//freopen("a.txt","r",stdin);
//freopen("b.txt","w",stdout);
int m,a,b,x,y,v;
char opt[5];
n = read();m = read();
init();
for(int i = 1;i<=n;++i) arr[i] = read();
for(int i = 1;i<n;++i)
{
a = read();b=read();
add(a,b);add(b,a);
}
dfs1(1,0);
dfs2(1,1);
build(1,1,n);
for(int i = 1;i<=m;++i)
{
scanf("%s",opt);
if(opt[0]=='C')
{
x = read();y=read();v=read();
updates(x,y,v);
}
else
{
x = read();y = read();
int LCA = lca(x,y);
int ans = sum(x,LCA)+sum(y,LCA);
printf("%d\n",ans-1);
}
}
//fclose(stdin);
//fclose(stdout);
//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
return 0;
}
![CF1394C 计算几何[图]](data:image/gif;base64,R0lGODlhAQABAIAAAP///wAAACwAAAAAAQABAAACAkQBADs=)