"Bzoj" 3319: Black and White tree (and check set + special tricks/-tree chain split + line tree)

http://www.lydsy.com/JudgeOnline/problem.php?id=3319

I think it is a template problem to review the next hld ....... ...........

Then nlg^2n was taken by the tle ......... .............

Then look at the Qaq,,, this ...

God-Problem practice ... I'll write again later ...

HLD of Tle:

#include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream > #include <algorithm> #include <queue> #include <set> #include <map>using namespace std; typedef long Long LL; #define REP (i, n) for (int i=0; i< (n); ++i) #define FOR1 (i,a,n) for (int i= (a); i<= (n); ++i) #define For2 (i,a,n) for (int i= (a);i< (n), ++i) #define FOR3 (i,a,n) for (int i= (a); i>= (n); i.) #define FOR4 (i,a,n) for (int i= ( a);i> (n); i) #define CC (i,a) memset (i,a,sizeof (i)) #define READ (a) a=getint () #define PRINT (a) printf ("%d", a) # Define DBG (x) cout << (#x) << "=" << (x) << endl#define error (x) (! x) puts ("error"): 0) #define RDM (x, i) for (int i=ihead[x]; i; i=e[i].next) inline const int Getint () {int r=0, k=1; Char c=g Etchar (); for (; c< ' 0 ' | | C> ' 9 '; C=getchar ()) if (c== '-') k=-1; for (; c>= ' 0 ' &&c<= ' 9 '; C=getchar ()) r=r*10+c-' 0 '; return k*r; }const int N=1e6+10;int ihead[n], cnt;struct dat {int Next, to, ID }e[n<<1];void Add (int u, int v, int id) {e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; e[cnt].id=id;e[++cnt]. NEXT=IHEAD[V]; ihead[v]=cnt; E[cnt].to=u; E[cnt].id=id;} int n, Son[n], top[n], fa[n], sz[n], root, id[n], tot, ed[n], dep[n];void dfs1 (int x, int f) {sz[x]=1; int y; fa[x]=f; dep [X]=DEP[F]+1;RDM (x, i) if (f!= (y=e[i].to)) {dfs1 (y, x); Sz[x]+=sz[y];if (Sz[y]>sz[son[x]]) son[x]=y;}} void dfs2 (int x, int tp) {TOP[X]=TP; Id[x]=++tot;if (Son[x]) DFS2 (son[x], TP), RDM (x, i) if (E[i].to!=fa[x] && e[i].t O!=SON[X]) dfs2 (e[i].to, e[i].to);} void geted (int x, int t) {ED[ID[X]]=T;RDM (x, i) if (E[i].to!=fa[x]) geted (e[i].to, e[i].id);} #define LC x<<1#define RC X<<1|1#define Lson L, Mid, Lc#define Rson mid+1, R, Rc#define mid (L+r) >>1stru CT T {int mx; bool tag;} T[n<<2];void upd (int x, int r) {T[x].tag=1;t[x].mx=max (t[x].mx, R);} void pushdown (int x, int mid, int r) {if (T[x].tag) t[x].tag=0, UPD (LC, mid), UPD (RC, r);} void pushup (int x) {T[x].mx=max (T[LC].mx, t[rc].mx); }void Update (int L, int R, int l=1, int r=tot, int x=1) {if (l<=l && r<=r) {upd (x, R); return;} int Mid=mid;pushdown (x, Mid, R), if (l<=mid) Update (L, R, Lson), if (mid<r) Update (L, R, Rson);p ushup (x);} int ask (int L, int R, int l=1, int r=tot, int x=1) {if (l<=l && r<=r) return t[x].mx;int Mid=mid, ret=0;pushd Own (x, Mid, R), if (L<=mid) Ret=max (ret, Ask (L, R, Lson)), if (Mid<r) Ret=max (ret, Ask (L, R, Rson)); return ret;} void ask (int x) {int fx=top[x], pos=0;while (fx!=root) {pos=ask (Id[fx], id[x]), if (pos!=0) BREAK;X=FA[FX]; fx=top[x];} if (pos==0 && x!=root) pos=ask (Id[top[x], id[x]);p rintf ("%d\n", Ed[pos]);} void change (int x, int y) {int fx=top[x], fy=top[y];while (fx!=fy) {if (Dep[fx]<dep[fy]) {swap (FX, FY), swap (x, y);} Update (ID[FX], id[x]); X=FA[FX]; FX=TOP[X];} if (x==y) return;if (Dep[x]<dep[y]) swap (x, y); Update (id[y]+1, id[x]);} int main () {read (n); int m=getint (); For1 (i, 1, n-1) {int u=getint (), V=getint (), Add (U, V, i);}ROOT=1;DFS1 (root, 0); DFS2 (root, root); geted (root, 0), while (m--) {int Cc=getint (), if (cc==1) Ask (Getint ()), else {int u=getint (), V=getint (), Change (U, v);}} return 0;}

　　

Description

Given a tree, the color of the edges is black or white, and the original is all white. Maintain two operations:


1. Query the label of the first black edge on the root path.
2. Set the color of all edges on the U to v path to black.

Notice: The root node of this tree is 1

Input

The first row of two number n,m represents the number of points and operands, respectively.
Next n? 1 lines, 2 numbers per line u,v. Represents a U to v edge.
Next m line, each behaves in the following format:


1 v indicates the first operation


2 v U represents the second operation

Output

For each query, output the corresponding answer. If not present, output 0.

Sample INPUT5 4
1 2
1 3
2 4
2 5
1 2
2 2 3
1 3
1 4

Sample Output0
2
1
HINT

For 100% data: n,m<=10^6

Source

"Bzoj" 3319: Black and White tree (and check set + special tricks/-tree chain split + line tree)