Codeforces Round #326 (Div. 1)-C. Duff in the Army tree multiplication algorithm

Test instructions: A number of n points, M personal live at some point in which each person's label 1-m, ask U-v path on the label before a person, and output designator, a < 10.

Practice, the use of multiplication, id[j][i] said I to I 2^j ancestors of the first 10 people, then each query direct maintenance is good, the details are many, just beginning do not know how to ask Id[j][i].

Here the 2^j is divided into two parts, the former 2^ (j-1) and the rear 2^ (j-1), and then the maintenance of the recursion.

Feel the tree chain can also be done, do not know if it will be tle, tree chain subdivision of the word segment tree each point maintenance 10 values, each merge on the line.

#include <bits/stdc++.h>using namespacestd;Const intMAXN = 1e5 +5;Const intMAXDEP = -;intPAR[MAXDEP][MAXN], DEP[MAXN], N;vector<int>ID[MAXDEP][MAXN], Cit[maxn];vector<int>G[MAXN];voidinit () { for(inti =0; i < MAXN; i++) {g[i].clear ();        Cit[i].clear ();  for(intj =0; J < MAXDEP; J + +) {id[j][i].clear (); }} memset (Par,-1,sizeof(PAR));}voidUpdate (Vector <int> &v1, Vector <int> &v2) {     for(intx:v2)    {v1.push_back (x);    } sort (V1.begin (), V1.end ());    V1.erase (Unique (V1.begin (), V1.end ()), V1.end ());  while(V1.size () >Ten) {v1.pop_back (); }}voidDfsintUintfather) {par[0][u] =father; Dep[u]= Dep[father] +1;  for(inti = u; I <= u; i++) {Update (id[0][i], cit[par[0][i]]); Update (id[0][i], cit[i]);  for(intj =0; J +1< MAXDEP; J + +) {            if(~Par[j][i]) {Par[j+1][i] =Par[j][par[j][i]]; Update (Id[j+1][i], cit[i]); Update (Id[j+1][i], id[j][i]); Update (Id[j+1][i], Id[j][par[j][i]]); } Else{par[j+1][i] =-1; }        }    }     for(intV:g[u]) {        if(V! =father)        {DFS (V, u); }    }}intLcaintUintv) {if(Dep[u] >Dep[v])    {Swap (U, v); }     for(intK =0; K < MAXDEP; k++) {        if((Dep[v]-dep[u]) >> K &1) {v=Par[k][v]; }    }    if(U = =v) {returnu; }     for(inti = maxdep-1; I >=0; i--) {        if(Par[i][u]! =Par[i][v]) {u=Par[i][u]; V=Par[i][v]; }    }    returnpar[0][u];}intAnc;vector<int> Solve (intUintv) {vector<int>Res; if(U = =v) {update (res, cit[u]); }     for(inti = maxdep-1; I >=0; i--) {        if(~par[i][u] && Dep[par[i][u]] >=Dep[anc])            {Update (res, id[i][u]); U=Par[i][u]; }    }     for(inti = maxdep-1; I >=0; i--) {        if(~par[i][v] && Dep[par[i][v]] >=Dep[anc])            {Update (res, id[i][v]); V=Par[i][v]; }} vector<int>EMP;    Update (res, EMP); returnRes;}intMain () {#ifndef Online_judge freopen ("In.txt","R", stdin);#endif    intm, q;  while(~SCANF ("%d%d%d", &n, &m, &q)) {init ();  for(inti =0; I < n1; i++) {            intu, v; scanf ("%d%d", &u, &v);            G[u].push_back (v);        G[v].push_back (U); }         for(inti =0; I < m; i++) {            intC; scanf ("%d", &c); if(Cit[c].size () <Ten) {Cit[c].push_back (i+1); id[0][c].push_back (i+1); }} DFS (1,0);  while(q--) {            intu, V, A; scanf ("%d%d%d", &u, &v, &a); ANC=LCA (U, v); Auto Res=Solve (U, v); inttot = min ((int) Res.size (), a); printf ("%d%c"Tot"\ n"[!tot]);  for(inti =0; I < min ((int) Res.size (), a); i++) {printf ("%d%c", Res[i],"\ n"[i+1==min ((int) Res.size (), a)]); }        }    }    return 0;}

Codeforces Round #326 (Div. 1)-C. Duff in the Army tree multiplication algorithm