Bzoj_3653_ Laughing _ Tree Array

Bzoj_3653_ Laughing _ Tree Array

Description set T for a tree with a root tree, we do the following definition:? Set A and B as two different nodes in T. If a is the ancestor of B, then it is said that "a than B does not know where the gaoming went". Set A and B as two different nodes in T. If the distance between A and B does not exceed a given constant x in the tree, then "A and B" is called "laughing". Given a root tree t of n nodes, the nodes are numbered 1 to n and the root node is node # 1th. You need to answer Q a question, ask the given two integers p and K, and ask how many ordered triples (A;B;C) satisfy: 1. A, B and C are three different points in T, and A is P node; 2. A and B are better than C do not know where to go; 3. A and B are laughing. The constant in the laughing here is the given K. The first line of input contains two positive integers n and q, each representing the number of points and queries that have a root tree. Next n-1 lines, each line describes an edge on a tree. Each row contains two integers u and V, which represents an edge between the node U and v. Next the Q line, each line describes an operation. Line I contains two integers representing the p and K of the I inquiry respectively. 1<=p<=n1<=k<=nn<=300000q<=300000output

Output Q line, each line corresponding to a query, representative of the answer to the question.

Sample Input5 3
0 S
1 3
2 4
4 5
2 2
4 1
2 3Sample Output3
1
3 b There are two possible positions, the ancestor of a and the subtree of a. If B is an ancestor of a, then C can only select one of the a subtree except a. If B is a point in the subtree of a, C can only select the B subtree except one of the B sub-trees. The problem is then transformed into a subtree with a distance less than or equal to K in the tree. Two restrictions: the subtree within and the depth is less than or equal to a value. The Dfs sequence position and depth of each point is considered as two coordinates, and transformed into two-dimensional points, which are solved with a tree-like array. Code:

#include <stdio.h> #include <string.h> #include <algorithm>using namespace std; #define N 300050typedef Long Long ll;int head[n],to[n<<1],nxt[n<<1],cnt,n,m;int dep[n],siz[n],dfn[n],s[n],son[n]; ll c[n],ans[n];inline void Add (int u,int v) {to[++cnt]=v; nxt[cnt]=head[u]; head[u]=cnt;} void fix (int x,int v) {for (;x<=n;x+=x& (x)) C[x]+=v;} ll inq (int x) {ll re=0; for (;x;x-=x& (x)) re+=c[x]; return re;} struct Qaq {int p,d,id,opt;}     A[n<<1];bool CMP (const QAQ &x,const qaq &y) {if (X.P==Y.P) return x.opt<y.opt; return X.P&LT;Y.P;} void Dfs (int x,int y) {int i; S[++s[0]]=x; Dfn[x]=s[0]; dep[x]=dep[y]+1;    Siz[x]=1;    for (I=head[x];i;i=nxt[i]) if (to[i]!=y) {DFS (to[i],x); Siz[x]+=siz[to[i]]; } Son[x]=s[0];}    int main () {scanf ("%d%d", &n,&m);    int i,x,y;    for (i=1;i<n;i++) {scanf ("%d%d", &x,&y); Add (x, y); add (y,x);    } dfs (1,0);    int tot=0;        for (i=1;i<=m;i++) {scanf ("%d%d", &x,&y);        Ans[i]=1ll*min (y,dep[x]-1) * (siz[x]-1);        int depp=min (DEP[X]+Y,N); A[++TOT].P=DFN[X]; A[tot].opt=-1; A[tot].d=depp;        A[tot].id=i; A[++TOT].P=SON[X];  A[tot].opt=1; A[tot].d=depp;    A[tot].id=i;    } sort (a+1,a+tot+1,cmp);    int now=0;        for (i=1;i<=tot;i++) {while (NOW&LT;=N&AMP;&AMP;NOW&LT;A[I].P) Now++,fix (dep[s[now]],siz[s[now]]-1);    Ans[a[i].id]+=a[i].opt*inq (A[I].D); } for (i=1;i<=m;i++) printf ("%lld\n", Ans[i]);}

Bzoj_3653_ Laughing _ Tree Array