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Hysbz 2243 staining (tree chain splitting)
2022-07-05 21:44:00 【Full stack programmer webmaster】
Hello everyone , I meet you again , I'm the king of the whole stack
The mood of doing the topic : This question is very thoughtful . Debugging code debugging for a long time . Write segment tree interval merging for the first time .
Their thinking :
The tree chain splits + Segment tree interval merging
The endpoint of the line segment tree records the color of the left and right intervals . Number of colors . When merging, use the idea of interval merging .
One more thing to note . When changing from one chain to another, it is necessary to infer whether the current node is the same color as the parent node .
Code :
#include<iostream>
#include<sstream>
#include<map>
#include<cmath>
#include<fstream>
#include<queue>
#include<vector>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<stack>
#include<bitset>
#include<ctime>
#include<string>
#include<cctype>
#include<iomanip>
#include<algorithm>
using namespace std ;
#define INT __int64
#define L(x) (x * 2)
#define R(x) (x * 2 + 1)
const int INF = 0x3f3f3f3f ;
const double esp = 0.0000000001 ;
const double PI = acos(-1.0) ;
const int mod = 1e9 + 7 ;
const int MY = 1400 + 5 ;
const int MX = 100000 + 5 ;
int n ,m ,idx ,num ;
int head[MX] ,ti[MX] ,top[MX] ,dep[MX] ,siz[MX] ,son[MX] ,father[MX] ,g[MX] ;
struct Edge
{
int v ,next ;
}E[MX*2] ;
void addedge(int u ,int v)
{
E[num].v = v ; E[num].next = head[u] ; head[u] = num++ ;
E[num].v = u ; E[num].next = head[v] ; head[v] = num++ ;
}
void dfs_find(int u ,int fa)
{
dep[u] = dep[fa] + 1 ;
siz[u] = 1 ;
son[u] = 0 ;
father[u] = fa ;
for(int i = head[u] ;i != -1 ;i = E[i].next)
{
int v = E[i].v ;
if(v == fa) continue ;
dfs_find(v ,u) ;
siz[u] += siz[v] ;
if(siz[son[u]] < siz[v]) son[u] = v ;
}
}
void dfs_time(int u ,int fa)
{
ti[u] = idx++ ;
top[u] = fa ;
if(son[u]) dfs_time(son[u] ,top[u]) ;
for(int i = head[u] ;i != -1 ;i = E[i].next)
{
int v = E[i].v ;
if(v == father[u] || v == son[u]) continue ;
dfs_time(v ,v) ;
}
}
struct node
{
int le ,rt ,lc ,rc ,num ,add ;
}T[MX*4] ;
void build(int x ,int le ,int rt)
{
T[x].le = le ; T[x].rt = rt ;
T[x].lc = T[x].rc = T[x].add = -1 ;
T[x].num = 0 ;
if(le == rt) return ;
int Mid = (le + rt)>>1 ;
build(L(x) ,le ,Mid) ;
build(R(x) ,Mid + 1 ,rt) ;
}
void push_down(int x)
{
if(T[x].add != -1)
{
// Left
T[L(x)].lc = T[L(x)].rc = T[L(x)].add = T[x].add ; T[L(x)].num = 1 ;
// Right
T[R(x)].lc = T[R(x)].rc = T[R(x)].add = T[x].add ; T[R(x)].num = 1 ;
T[x].add = -1 ;
}
}
void push_up(int x)
{
T[x].num = T[L(x)].num + T[R(x)].num ;
if(T[L(x)].rc == T[R(x)].lc)
T[x].num -- ;
T[x].lc = T[L(x)].lc ; T[x].rc = T[R(x)].rc ;
}
void update(int x ,int le ,int rt ,int w) // Update an interval
{
if(T[x].le == le && T[x].rt == rt)
{
T[x].add = w ;
T[x].lc = w ; T[x].rc = w ; T[x].num = 1 ;
return ;
}
push_down(x) ;
int Mid = (T[x].le + T[x].rt)>>1 ;
if(le > Mid) update(R(x) ,le ,rt ,w) ;
else if(rt <= Mid) update(L(x) ,le ,rt ,w) ;
else
{
update(L(x) ,le ,Mid ,w) ;
update(R(x) ,Mid+1 ,rt ,w) ;
}
push_up(x) ;
}
int Query(int x ,int le ,int rt)
{
if(T[x].le == le && T[x].rt == rt)
return T[x].num ;
int Mid = (T[x].le + T[x].rt)>>1 ;
push_down(x) ;
if(le > Mid) return Query(R(x) ,le ,rt) ;
else if(rt <= Mid) return Query(L(x) ,le ,rt) ;
else
{
int mx = 0 ;
if(T[L(x)].rc == T[R(x)].lc)
mx = -1 ;
return Query(L(x) ,le ,Mid) + Query(R(x) ,Mid+1 ,rt) + mx ;
}
push_up(x) ;
}
int Querynode(int x ,int pos)
{
if(T[x].le == T[x].rt)
return T[x].lc ;
int Mid = (T[x].le + T[x].rt)>>1 ;
push_down(x) ;
if(pos <= Mid)
return Querynode(L(x) ,pos) ;
else return Querynode(R(x) ,pos) ;
push_up(x) ;
}
void LCAC(int u ,int v ,int w)
{
while(top[u] != top[v])
{
if(dep[top[u]] < dep[top[v]])
swap(u ,v) ;
update(1 ,ti[top[u]] ,ti[u] ,w) ;
u = father[top[u]] ;
}
if(dep[u] > dep[v])
swap(u ,v) ;
update(1 ,ti[u] ,ti[v] ,w) ;
}
int LCAQ(int u ,int v)
{
int ans = 0 ;
while(top[u] != top[v])
{
if(dep[top[u]] < dep[top[v]])
swap(u ,v) ;
ans += Query(1 ,ti[top[u]] ,ti[u]) ;
if(Querynode(1 ,ti[top[u]]) == Querynode(1 ,ti[father[top[u]]]))
ans-- ;
u = father[top[u]] ;
}
if(dep[u] > dep[v])
swap(u ,v) ;
ans += Query(1 ,ti[u] ,ti[v]) ;
return ans ;
}
int main()
{
while(~scanf("%d%d" ,&n ,&m))
{
int u ,v ,w ;
num = 0 ;
memset(head ,-1 ,sizeof(head)) ;
for(int i = 1 ;i <= n ; ++i)
scanf("%d" ,&g[i]) ;
for(int i = 1 ;i < n ; ++i)
{
scanf("%d%d" ,&u ,&v) ;
addedge(u ,v) ;
}
dep[1] = siz[0] = 0 ;
dfs_find(1 ,1) ;
idx = 1 ;
dfs_time(1 ,1) ;
build(1 ,1 ,n) ;
for(int i = 1 ;i <= n ; ++i)
update(1 ,ti[i] ,ti[i] ,g[i]) ;
char s[5] ;
for(int i = 0 ;i < m ; ++i)
{
scanf("%s" ,s) ;
scanf("%d%d" ,&u ,&v) ;
if(s[0] == 'C')
{
scanf("%d" ,&w) ;
LCAC(u ,v ,w) ;
}
else if(s[0] == 'Q')
printf("%d\n" ,LCAQ(u ,v)) ;
}
}
return 0 ;
}Copyright notice : This article is an original blog article , Blog , Without consent , Shall not be reproduced .
Publisher : Full stack programmer stack length , Reprint please indicate the source :https://javaforall.cn/117566.html Link to the original text :https://javaforall.cn
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