【poj 1741】Tree 【树分治 点分治入门题】

Tree

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 18550 Accepted: 6069

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

树分治之点分治算法

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<cmath>#include<vector>using namespace std;int n,k;#define maxn 10005typedef pair<int,int> pii;#define mk(a,b) make_pair(a,b)vector<pii>g[maxn<<2];int rt,s[maxn],f[maxn],vis[maxn],d[maxn],ans,size,q[maxn];/********* s[i] -> i的size（算上本身子树的节点数） * f[i] -> i的子节点中s[v]最大的值. * 性质：f[i]最小的点是树的重心 *********///getroot() => 计算当前树的重心 void getroot(int u,int fa){s[u] = 1;f[u] = 0;for(int i=0;i<g[u].size();i++){int v = g[u][i].first;if(v==fa||vis[v])continue; //考虑过 getroot(v,u); s[u]+=s[v];f[u] = max(f[u],s[v]);}f[u] = max(f[u],size-f[u]); if(f[u]<f[rt])rt = u; }//dfs() => 计算深度void dfs(int u,int fa){s[u] = 1;q[++(*q)]=d[u];for(int i=0;i<g[u].size();i++){int v = g[u][i].first;if(v==fa||vis[v])continue;d[v] = d[u] + g[u][i].second;dfs(v,u);s[u]+=s[v];}} int calc(int u,int deep){d[u] = deep, *q = 0;dfs(u,0);sort(q+1,q+1+(*q));int l = 1,r = *q,ret = 0;while(l<r){if(q[l]+q[r]<=k)ret+=r-l++;else r--;}return ret;}void solve(int u){vis[u] = 1;ans += calc(u,0);//加上 d[i]+d[j]<=k的(i,j)对数for(int i=0;i<g[u].size();i++){int v = g[u][i].first;if(vis[v])continue;ans-=calc(v,g[u][i].second);f[0] = size = s[v];getroot(v,rt = 0);solve(rt);}}int main(void){int a,b,c;while(scanf("%d%d",&n,&k)&&(n+k)){for(int i=0;i<=n;i++)g[i].clear();memset(vis,0,sizeof(vis));for(int i=1;i<n;i++){scanf("%d%d%d",&a,&b,&c);g[a].push_back(mk(b,c));g[b].push_back(mk(a,c));}f[0] = size = n;//初始化极大值 getroot(1,rt=0);ans = 0;solve(rt);printf("%d\n",ans);}return 0;}