CodeForces161D: Distance in Tree
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A tree is a connected graph that doesn‘t contain any cycles.
The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.
You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.
Input
The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.
Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.
Output
Print a single integer — the number of distinct pairs of the tree‘s vertices which have a distance of exactly k between them.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
5 2
1 2
2 3
3 4
2 5
4
5 3
1 2
2 3
3 4
4 5
2
题意:给定一棵树,问树上有多少个点对距离是K(K<=500)。
思路:明显的基础分治题,分别累计经过根节点的距离为K的点对。说他基础是以为既不需要排序,也不需要去重。复杂度O(NlogN*K)
#include<bits/stdc++.h> using namespace std; const int maxn=100050; int Laxt[maxn],Next[maxn],To[maxn]; int sz[maxn],son[maxn],root,cnt,N,K,ans,sn; //sn,小树的大小。 int num[550],tnum[550],vis[maxn]; void read(int &x){ x=0; char c=getchar(); while(c>‘9‘||c<‘0‘) c=getchar(); while(c>=‘0‘&&c<=‘9‘) x=(x<<3)+(x<<1)+c-‘0‘,c=getchar(); } void add(int u,int v) { Next[++cnt]=Laxt[u]; Laxt[u]=cnt; To[cnt]=v; } void getroot(int u,int fa) { sz[u]=1; son[u]=0; for(int i=Laxt[u];i;i=Next[i]){ int v=To[i]; if(v==fa||vis[v]) continue; getroot(v,u); sz[u]+=sz[v]; son[u]=max(son[u],sz[v]); } son[u]=max(sz[u],sn-son[u]); if(root==0||son[root]>son[u]) root=u; } void getdep(int u,int fa,int dis) { if(K>=dis) ans+=num[K-dis],tnum[dis]++; if(K==dis) return ; for(int i=Laxt[u];i;i=Next[i]) if(To[i]!=fa&&!vis[To[i]]) getdep(To[i],u,dis+1); } void dfs(int u) { vis[u]=1; for(int i=1;i<=K;i++) num[i]=0; num[0]=1; for(int i=Laxt[u];i;i=Next[i]){ //暴力部分 if(vis[To[i]]) continue; for(int j=0;j<=K;j++) tnum[j]=0; getdep(To[i],0,1); for(int j=0;j<=K;j++) num[j]+=tnum[j]; } for(int i=Laxt[u];i;i=Next[i]){ //以大化小,分治部分 if(vis[To[i]]) continue; sn=sz[To[i]]; root=0; getroot(To[i],u); dfs(root); } } int main() { int u,v; scanf("%d%d",&N,&K); for(int i=1;i<N;i++){ read(u) ; read(v) ; add(u,v); add(v,u); } root=0; sn=N; getroot(1,0); dfs(root); printf("%d\n",ans); return 0; }
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