"Codeforces" E. Xenia and Tree (sub-block + LCA)

For two operations, for Operation 1, save the points to be modified each time, until the sqrt (m) point is saved, and the diagram is rebuilt once (the distance from each point to the nearest staining point is re-recorded)

For Operation 2, use the LCA to find the current point (i.e., the point that has not been updated to the graph) and the distance between the point and the point in the current graph and the nearest stain.

It's a good thing to say about LCA = =! Fill in these days.

The puzzle is here: http://codeforces.com/blog/entry/8800

#include <cmath> #include <queue> #include <cstdio> #include <vector> #include <cstring># include<algorithm>using namespace Std;const int maxn = 100005;const int INF = 9999999;int N,m;vector<int>g[ma Xn];int pa[maxn][25];int dist[maxn];int deep[maxn];int vis[maxn];//--------------------    LCA---------------------------//multiplication method Lac for nearest public ancestor void Dfs (int pos,int d) {Deep[pos] = D;    int Size = G[pos].size (); for (int i = 0; i < Size; i++) DFS (g[pos][i],d + 1);} void Lca_init () {for (int j = 1; J <=; J + +) for (int i = 1; I <= n; i++) if (pa[i][j-1]! =-1 ) Pa[i][j] = pa[pa[i][j-1]][j-1];}    int LCA (int a,int b) {int J;    if (Deep[a] < deep[b]) swap (A, b);            Speak AB at the same depth for (int j = k; j >= 0; j--) {if (pa[a][j]! = 1 && deep[pa[a][j]] >= deep[b])    A = Pa[a][j];    }//if (a = = b) return b; Multiplication method for (int j = 0; J >=; j--) {if (pa[a][j]! =-1 && pa[a][j]! = Pa[b][j]) {a = Pa[a][j];        b = Pa[b][j]; }} return pa[a][0];}    --------------------------------------------------void BFs () {queue<int>q;    int HAVE[MAXN] = {0};            for (int i = 1; I <= n; i++) if (Vis[i]) {//If staining q.push (i);            Have[i] = 1;        Dist[i] = 0;        } while (!q.empty ()) {Int. now = Q.front (); Q.pop ();        int Size = G[now].size ();            for (int i = 0; i < Size; i++) if (!have[g[now][i]]) {Have[g[now][i]] = 1;            Dist[g[now][i]] = Dist[now] + 1;        Q.push (G[now][i]);            } if (pa[now][0]! =-1 &&!have[pa[now][0]]) {have[pa[now][0]] = 1;            Dist[pa[now][0]] = Dist[now] + 1;        Q.push (Pa[now][0]); }}}//--------------------------------------------------vector<int>vv;void init () {for (int i = 1; I <= n; i    + +) G[i].clear ();    Vv.clear (); memset (vis,-1,sizeof (Vis)); memset (Pa,-1,sizeof (PA));}        int main () {while (scanf ("%d%d", &n,&m)! = EOF) {init ();            for (int i = 0; i < n-1; i++) {int x, y;            scanf ("%d%d", &x,&y);            G[x].push_back (y);            Pa[y][0] = x;        Vis[y] = 0;            }//Find the root node and get the deep array for (int i = 1; I <= n; i++) if (vis[i] = = 1) {pa[i][0] = 1;            Vis[i] = 0;            VIS[1] = 1;            DFS (i,0);        Break        } lca_init ();        BFS ();        int calc = sqrt (m) + 1;            for (int i = 0; i < m; i++) {int op,v;            scanf ("%d%d", &op,&v);                if (op = = 1) {vv.push_back (v);                int S = Vv.size ();                    if (s = = Calc) {for (int i = 0; i < S; i++) vis[vv[i]] = 1;                    BFS ();                Vv.clear (); }} else{int S = Vv.sizE ();                int ans = dist[v];                    for (int i = 0; i < S; i++) {int c = LCA (V,vv[i]);                    int d = (Deep[v]-deep[c]) + (Deep[vv[i]]-deep[c]);                ans = min (ans,d);            } printf ("%d\n", ans); }}} return 0;}

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"Codeforces" E. Xenia and Tree (sub-block + LCA)