BZOJ1103: [POI2007] Metropolitan Meg

Source: Internet
Author: User

The DFS sequence.

Use L[u] and r[u] to indicate the time to enter u and go out U. Use a tree-like array to maintain prefixes and l[u] minus 1 at 1,r[u].

The question is the distance from the root to yourself, which is equivalent to the prefix of L[u]. Why is it?

If a node V is on the root-to-u path, it will add 1 to the l[v] place. If not, if the numbers are less than u,l[v] and r[v] are eliminated.

L[V] greater than l[u] if it is greater than U. W u operation is to L[u] minus 1, will r[u] place plus 1. This will not affect the path length.

The ingenious application of the DFS sequence.

#include <cstdio>#include<algorithm>#include<cstring>using namespacestd;Const intMAXN =250000+Ten;Const intMAXM =500000+Ten;structBIT {inta[maxn<<1],n; InlineintLowbit (intx) {return(x& (-x)); }        voidAddintXintv) { for(; x<=n;x+=lowbit (x)) a[x]+=v; }        intSumintx) {intres=0;  for(; x;x-=lowbit (x)) res+=A[x]; returnRes; }        voidInitintm) {memset (A,0,sizeof(a)); N=m; }} bit;intG[maxn],v[maxm],next[maxm],eid;intFA[MAXN],L[MAXN],R[MAXN],S[MAXN],SP,DFN;intn,m;voidAddedge (intAintb) {V[eid]=b; Next[eid]=g[a]; g[a]=eid++; V[eid]=a; NEXT[EID]=G[B]; g[b]=eid++;}voidDfs () {s[++sp]=1;  while(sp) {intu=s[sp--]; if(!L[u]) {L[u]=++DFN; s[++sp]=T;  for(intI=g[u];~i;i=next[i])if(v[i]!=Fa[u]) {Fa[v[i]]=u; s[++sp]=V[i]; }        }        Elser[u]=++DFN; }}intMain () {scanf ("%d",&N); Memset (g,-1,sizeof(g)); Bit.init (n*2);  for(intI=1, a,b;i<n;i++) {scanf ("%d%d",&a,&b);    Addedge (A, b);        } dfs ();  for(intI=2; i<=n;i++) {Bit.add (l[i],1); Bit.add (R[i],-1); }    Charop[Ten]; scanf ("%d", &m); m+= (n1);  for(intI=1, a,b;i<=m;i++) {scanf ("%s%d",s,&a); if(s[0]=='A') {scanf ("%d",&C); if(a>b) Swap (A, a); Bit.add (L[b],-1); Bit.add (R[b],1); }        Elseprintf"%d\n", Bit.sum (L[a])); }    return 0; }

BZOJ1103: [POI2007] Metropolitan Meg

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