poj1741树分治

Tree

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 11912 Accepted: 3755

Description

Give a tree with n vertices,each edge has a length(positive integer less than 1001). 
Define dist(u,v)=The min distance between node u and v. 
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k. 
Write a program that will count how many pairs which are valid for a given tree. 

Input

The input contains several test cases. The first line of each test case contains two integers n, k. (n<=10000) The following n-1 lines each contains three integers u,v,l, which means there is an edge between node u and v of length l. 
The last test case is followed by two zeros. 

Output

For each test case output the answer on a single line.

Sample Input

5 41 2 31 3 11 4 23 5 10 0

Sample Output

8

Source

LouTiancheng@POJ

题意：给定一棵树和距离k，求距离小于k节点对的个数

解法：在树上做分治。按照重心分解，分成在同一颗子树上和不同子树上两种情况讨论，不过要注意去重

代码：

////  main.cpp//  poj1741////  Created by 蘇與軒 on 15/2/9.//  Copyright (c) 2015年 蘇與軒. All rights reserved.//#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <cmath>#include <cstdlib>#include <string>#include <map>#include <set>#include <algorithm>#include <functional>#define rep(i,a,b) for (int i=a;i<((b)+1);i++)#define Rep(i,a,b) for (int i=a;i>=b;i--)#define foreach(e,x) for (__typeof(x.begin()) e=x.begin();e!=x.end();e++)#define mid ((l+r)>>1)#define lson (k<<1)#define rson (k<<1|1)#define MEM(a,x) memset(a,x,sizeof a)using namespace std;const int N=2000050;typedef pair<int, int> pii;typedef long long ll;int pnt[N],head[N],nxt[N],dist[N],n,k,cnt,ans,size,f[N],tsize[N],root,d[N];bool vis[N];vector<int> dis;void addedge(int u,int v,int c) {    pnt[cnt]=v;nxt[cnt]=head[u];dist[cnt]=c;head[u]=cnt++;    pnt[cnt]=u;nxt[cnt]=head[v];dist[cnt]=c;head[v]=cnt++;}void subtree(int u,int p) {    int v;    tsize[u]=1;f[u]=0;    for (int i=head[u];i!=-1;i=nxt[i]) {        v=pnt[i];        if (v==p||vis[v]) continue;        subtree(v, u);        tsize[u]+=tsize[v];        f[u]=max(f[u],tsize[v]);    }    f[u]=max(f[u],size-f[u]);    if (f[u]<f[root])   root=u;}void getpath(int u,int p) {    dis.push_back(d[u]);    int v;    tsize[u]=1;    for (int i=head[u];i!=-1;i=nxt[i]) {        v=pnt[i];        if (v!=p&&!vis[v]) {            d[v]=d[u]+dist[i];            getpath(v, u);            tsize[u]+=tsize[v];        }    }}int cal(int u,int init) {    dis.clear();d[u]=init;    getpath(u,0);    sort(dis.begin(),dis.end());    int ret=0;    for (int l=0,r=(int)dis.size()-1;l<r;)        if (dis[l]+dis[r]<=k)   ret+=r-l++;        else r--;    return ret;}void solve(int u) {    vis[u]=true;    int v;    for (int i=head[u];i!=-1;i=nxt[i]) {        v=pnt[i];        if (!vis[v]) {            f[0]=size=tsize[u];            ans-=cal(v,dist[i]);            subtree(v, root=0);            solve(root);        }    }    ans+=cal(u,0);    vis[u]=false;}int main(int argc, const char * argv[]) {    while (scanf("%d%d",&n,&k)!=EOF){        if ((n|k)==0) break;        MEM(head,-1);MEM(nxt,-1);        rep(i,1,n-1){            int u,v,c;            scanf("%d%d%d",&u,&v,&c);            addedge(u,v,c);        }        size=f[0]=n;        subtree(1,root=0);        ans=0;        solve(root);        printf("%d\n",ans);    }    return 0;}