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LCA using Sparse Table
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//LCA using sparse table | |
//Complexity: O(NlgN,lgN) | |
#define mx 100002 | |
int L[mx]; //লেভেল | |
int P[mx][22]; //স্পার্স টেবিল | |
int T[mx]; //প্যারেন্ট | |
vector<int>g[mx]; | |
void dfs(int from,int u,int dep) | |
{ | |
T[u]=from; | |
L[u]=dep; | |
for(int i=0;i<(int)g[u].size();i++) | |
{ | |
int v=g[u][i]; | |
if(v==from) continue; | |
dfs(u,v,dep+1); | |
} | |
} | |
int lca_query(int N, int p, int q) //N=নোড সংখ্যা | |
{ | |
int tmp, log, i; | |
if (L[p] < L[q]) | |
tmp = p, p = q, q = tmp; | |
log=1; | |
while(1) { | |
int next=log+1; | |
if((1<<next)>L[p])break; | |
log++; | |
} | |
for (i = log; i >= 0; i--) | |
if (L[p] - (1 << i) >= L[q]) | |
p = P[p][i]; | |
if (p == q) | |
return p; | |
for (i = log; i >= 0; i--) | |
if (P[p][i] != -1 && P[p][i] != P[q][i]) | |
p = P[p][i], q = P[q][i]; | |
return T[p]; | |
} | |
void lca_init(int N) | |
{ | |
memset (P,-1,sizeof(P)); //শুরুতে সবগুলো ঘরে -১ থাকবে | |
int i, j; | |
for (i = 0; i < N; i++) | |
P[i][0] = T[i]; | |
for (j = 1; 1 << j < N; j++) | |
for (i = 0; i < N; i++) | |
if (P[i][j - 1] != -1) | |
P[i][j] = P[P[i][j - 1]][j - 1]; | |
} | |
int main(void) { | |
g[0].pb(1); | |
g[0].pb(2); | |
g[2].pb(3); | |
g[2].pb(4); | |
dfs(0, 0, 0); | |
lca_init(5); | |
printf( "%d\n", lca_query(5,3,4) ); | |
return 0; | |
} |
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