Problem Link : http://www.spoj.com/problems/BTCODE_G/
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The solution idea of this problem is not so hard. This one can be solve with Heavy Light Decomposition. But as the time limit is very tight we need a lot of optimization. To solve this problem we need to convert the tree to an array using Euler tour or in-order traversal on the tree. Now there are 2 types of operation. Let consider each tree node as a special node which contain a 30 size array for storing it’s color information.
Now let a paint operation is 1 a b. Now add +1 to array index [b] of discovery_time(a)node and -1 to array index [b] of finishing_time[a) node.
For each query operation 2 a b we need to check for each color. Now at first determine the LCA of node a and b in the original given graph. Now for the i’th color we need to count the summation of array[i] from root node to discovery_time(a) node. Let this summation is x. Then we need count the summation of array[i] from root node to discovery_time(b) node. Let this summation is y. Here we add root to LCA node’s value twice but we don’t need this. So we need to subtract this from the sum. So we need count the summation of array[i] from root node to discovery_time(LCA) node. Let this is z. Now if we subtract 2*z then we will delete the LCA node’s value so wee need to add this value too.
So Total number of i’th color node in the path from a to b is
=x+y-(2*z)+ array[i] value of LCA node.
We can do this for each of 30 color and When we get a color which fill every node in the range a to b then we will print YES otherwise NO.
I have implemented this solution using segment tree but got TLE several time. Then I implement the same idea using BIT and got AC. I think the main idea behind the TLE in segment tree is wee need to copy the 30 size array element several time in segment tree. Which will take some time. As the time limit is very tight, the segment tree solution is inefficient.
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define pii pair <int,int>
#define pll pair <long long,long long>
#define sc scanf
#define pf printf
#define Pi 2*acos(0.0)
#define ms(a,b) memset(a, b, sizeof(a))
#define pb(a) push_back(a)
#define MP make_pair
#define db double
#define ll long long
#define EPS 10E-10
#define ff first
#define ss second
#define sqr(x) (x)*(x)
#define D(x) cerr<<#x " = "<<(x)<<endl
#define VI vector <int>
#define DBG pf("Hi\n")
#define MOD 1000000007
#define CIN ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a) (int)a.size()
#define sf(a) scanf("%d",&a)
#define sfl(a) scanf("%lld",&a)
#define sff(a,b) scanf("%d %d",&a,&b)
#define sffl(a,b) scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v) for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n) for(int i=0;i<n;i++)
#define loop1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define RREP(i,a,b) for(int i=a;i>=b;i--)
#define TEST_CASE(t) for(int z=1;z<=t;z++)
#define PRINT_CASE printf("Case %d: ",z)
#define LINE_PRINT_CASE printf("Case %d:\n",z)
#define CASE_PRINT cout<<"Case "<<z<<": "
#define all(a) a.begin(),a.end()
#define intlim 2147483648
#define infinity (1<<28)
#define ull unsigned long long
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a)*((b)/gcd(a,b)))
using namespace std;
//using namespace __gnu_pbds;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1}; // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1}; // Kings Move
//const int fx[]={-2, -2, -1, -1, 1, 1, 2, 2}; // Knights Move
//const int fy[]={-1, 1, -2, 2, -2, 2, -1, 1}; // Knights Move
/*------------------------------------------------*/
/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/
#define mx 200005
vector<int>g[mx];
pii times[mx];
int dfsnum;
int parent[mx],level[mx],node_cnt[2*mx];
void dfs(int u, int p)
{
times[u].ff=++dfsnum;
parent[u]=p;
for(int i=0; i<SZ(g[u]); i++)
{
int v=g[u][i];
if(v==p) continue;
level[v]=level[u]+1;
dfs(v,u);
}
times[u].ss=++dfsnum;
}
int dp[mx][21];
int func_lca(int idx, int p)
{
if(p==0)
{
return dp[idx][p]=parent[idx];
}
int &ret=dp[idx][p];
if(ret!=-1) return ret;
int u=func_lca(idx,p-1);
ret=func_lca(u,p-1);
return ret;
}
int lca_query(int p, int q)
{
if(level[q]>level[p]) swap(p,q);
for(int i=20;i>=0;i--)
{
int a=func_lca(p,i);
if(level[a]>=level[q])
p=a;
}
if(p==q) return p;
for(int i=20;i>=0;i--)
{
int a=func_lca(p,i);
int b=func_lca(q,i);
if(a!=b)
{
p=a;
q=b;
}
}
return parent[p];
}
int tree[31][2*mx];
void update(int id, int idx, int val)
{
for(;idx<=dfsnum && idx;idx+=idx&-idx)
tree[id][idx]+=val;
}
int query(int id, int idx)
{
int ret=0;
for(;idx;idx-=idx&-idx)
ret+=tree[id][idx];
return ret;
}
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
int n;
sf(n);
for(int i=1; i<n; i++)
{
int a,b;
sff(a,b);
g[a].pb(b);
g[b].pb(a);
}
dfs(0,0);
int qq;
sf(qq);
for(int i=0;i<n;i++)
{
node_cnt[times[i].ff]=1;
node_cnt[times[i].ss]=-1;
}
partial_sum(node_cnt,node_cnt+(2*mx),node_cnt);
ms(dp,-1);
while(qq--)
{
int a,b,c;
sfff(a,b,c);
if(a==1)
{
update(c,times[b].ff,1);
update(c,times[b].ss,-1);
}
else
{
if(times[b].ff>times.ff)
swap(b,c);
bool ans=0;
int lca=lca_query(b,c);
for(int i=1;i<=30;i++)
{
int x=query(i,times.ff);
int y=query(i,times[b].ff);
int z=query(i,times[lca].ff);
x=(x+y-(2*z));
x+=query(i,times[lca].ff)-query(i,times[lca].ff-1);
y=node_cnt[times[b].ff]+node_cnt[times.ff]-(2*node_cnt[times[lca].ff])+(node_cnt[times[lca].ff]-node_cnt[times[lca].ff-1]);
if(x)
{
if(x==y)
ans=1;
break;
}
}
if(ans==0)
printf("NO\n");
else
printf("YES\n");
}
}
return 0;
}