简单LCA:
求树上距离给定两个点a,b距离相等的点有多少个
先预处理出每个节点的孩子个数sum[x],求出a,b的LCA,根据深度就可以知道两个点的距离,距离为偶数的有解….
根据lca在a,b之间的位置不同分情况讨论:
设a与lca距离为 ha , b与lca距离为 hb
1:lca在a,b正中间既a,b分别属于lca的两个子树中, 结果为: n-sum[ a往上距离lca ha-1 的点] – sum[ b往上距离lca hb-1 的点]
sum[lca]-sum[ c往上距离lca hc-1 的点 ]
用倍增法求LCA O(logn) , 总时间复杂度 O(mlogn)
E. A and B and Lecture Rooms
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
A and B are preparing themselves for programming contests.
The University where A and B study is a set of rooms connected by corridors. Overall, the University hasnrooms connected byn-1corridors so that you can get from any room to any other one by moving along the corridors. The rooms are numbered from1ton.
Every day А and B write contests in some rooms of their university, and after each contest they gather together in the same room and discuss problems. A and B want the distance from the rooms where problems are discussed to the rooms where contests are written to be equal. The distance between two rooms is the number of edges on the shortest path between them.
As they write contests in new rooms every day, they asked you to help them find the number of possible rooms to discuss problems for each of the followingmdays.
Input
The first line contains integern(1≤n≤105) — the number of rooms in the University.
The nextn-1lines describe the corridors. Thei-th of these lines (1≤i≤n-1) contains two integersaiandbi(1≤ai,bi≤n), showing that thei-th corridor connects roomsaiandbi.
The next line contains integerm(1≤m≤105) — the number of queries.
Nextmlines describe the queries. Thej-th of these lines (1≤j≤m) contains two integersxjandyj(1≤xj,yj≤n) that means that on thej-th day A will write the contest in the roomxj, B will write in the roomyj.
Output
In thei-th (1≤i≤m) line print the number of rooms that are equidistant from the rooms where A and B write contest on thei-th day.
Sample test(s)
input
41 21 32 412 3
output
1
input
41 22 32 421 21 3
output
02
Note
in the first sample there is only one room at the same distance from rooms number 2 and 3 — room number 1.
/* ***********************************************Author:CKbossCreated Time :2015年03月05日 星期四 16时52分05秒File Name:E.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;const int maxn=100100;const int DEG=20;struct Edge{int to,next;}edge[maxn*2];int Adj[maxn],Size;void init_edge(){memset(Adj,-1,sizeof(Adj)); Size=0;}void Add_Edge(int u,int v){edge[Size].to=v;edge[Size].next=Adj[u];Adj[u]=Size++;}int fa[maxn][DEG];int deg[maxn];void BFS(int root){queue<int> q;deg[root]=0;fa[root][0]=root;q.push(root);while(!q.empty()){int u=q.front(); q.pop();for(int i=1;i<DEG;i++)fa[u][i]=fa[fa[u][i-1]][i-1];for(int i=Adj[u];~i;i=edge[i].next){int v=edge[i].to;if(v==fa[u][0]) continue;deg[v]=deg[u]+1;fa[v][0]=u;q.push(v);}}}int LCA(int u,int v){if(deg[u]>deg[v]) swap(u,v);int hu=deg[u],hv=deg[v];int tv=v,tu=u;for(int det=hv-hu,i=0;det;i++,det=det/2)if(det&1) tv=fa[tv][i];if(tv==tu) return tu;for(int i=DEG-1;i>=0;i–){if(fa[tu][i]==fa[tv][i]) continue;tu=fa[tu][i]; tv=fa[tv][i];}return fa[tu][0];}int n,m;int sum[maxn];int DFS(int u){int ret=1;for(int i=Adj[u];~i;i=edge[i].next){int v=edge[i].to;if(v==fa[u][0])continue;ret+=DFS(v);}return sum[u]=ret;}void solve(int x,int y){if(x==y) { printf("%d\n",n); return ; }int lca=LCA(x,y);int h1=deg[x]-deg[lca];int h2=deg[y]-deg[lca];if((h1+h2)%2==1) { puts("0"); return ; }if(h1==h2){int p1=x,p2=y;for(int det=(h1+h2)/2-1,i=0;det;i++,det/=2)if(det&1) p1=fa[p1][i];for(int det=(h1+h2)/2-1,i=0;det;i++,det/=2)if(det&1) p2=fa[p2][i];printf("%d\n",n-sum[p1]-sum[p2]);}else{if(deg[x]<deg[y]) swap(x,y);int p1=x,p2=x;for(int det=(h1+h2)/2,i=0;det;i++,det/=2)if(det&1) p1=fa[p1][i];for(int det=(h1+h2)/2-1,i=0;det;i++,det/=2)if(det&1) p2=fa[p2][i];int ans=sum[p1]-sum[p2];printf("%d\n",ans);}}int main(){//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);init_edge();scanf("%d",&n);for(int i=0;i<n-1;i++){int a,b;scanf("%d%d",&a,&b);Add_Edge(a,b); Add_Edge(b,a);}BFS(1); DFS(1);scanf("%d",&m);while(m–){int a,b;scanf("%d%d",&a,&b);solve(a,b);}return 0;}
，也有善意的提醒：何不去远方！昆明呀——赶一个花海;
