POJ1741(SummerTrainingDay08-G 树的点分治)

Tree

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 23380		Accepted: 7748

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 4
1 2 3
1 3 1
1 4 2
3 5 1
0 0

Sample Output

8

Source

LouTiancheng@POJ

 

 //2017-08-09
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <vector>

 using namespace std;

 const int N = ;
 int n, k, MAX, root, cnt, answer;

 //链式前向星
 int head[N], tot;
 struct Edge{
     int next, to, w;
 }edge[N<<];

 void add_edge(int u, int v, int w){
     edge[tot].w = w;
     edge[tot].to = v;
     edge[tot].next = head[u];
     head[u] = tot++;
 }

 int size[N];//size[i]表示以i为根的子树的大小，包括i。
 int maxson[N];//maxson[i]表示以i为根的子树的最大儿子的大小。
 int dis[N];//dis[i]表示i到根的距离。
 bool vis[N];//vis[i]用来标记i点是否被删除。

 void init(int n){
     answer = ;
     tot = ;
     memset(vis, , sizeof(vis));
     memset(head, -, sizeof(head));
 }

 //计算出子树的大小
 void dfs_size(int u, int fa){
     size[u] = ;
     maxson[u] = ;
     for(int i = head[u]; ~i; i = edge[i].next){
         int v = edge[i].to;
         if(vis[v] || v == fa)continue;
         dfs_size(v, u);
         size[u] += size[v];
         if(size[v] > maxson[u])
               maxson[u] = size[v];
     }
 }

 //找子树的重心。最大子树最小的点即为树的重心。
 void dfs_root(int r, int u, int fa){
     if(size[r] - size[u] > maxson[u])//size[r]-size[u]为u上面的树的尺寸
           maxson[u] = size[r] - size[u];
     if(maxson[u] < MAX){
         MAX = maxson[u];
         root = u;
     }
     for(int i = head[u]; ~i; i = edge[i].next){
         int v = edge[i].to;
         if(vis[v] || v == fa)continue;
         dfs_root(r, v, u);
     }
 }

 //计算出子树中每个点距离重心的距离
 void dfs_dis(int u, int d, int fa){
     dis[cnt++] = d;
     for(int i = head[u]; ~i; i = edge[i].next){
         int v = edge[i].to;
         if(vis[v] || v == fa)continue;
         dfs_dis(v, d+edge[i].w, u);
     }
 }

 //计算出以u为根的子树中距离和小于k的点对数
 int cal(int u, int d){
     int ans = ;
     cnt = ;
     dfs_dis(u, d, );
     sort(dis, dis+cnt);
     for(int i = , j = cnt-; i < j; i++){
         while(dis[i]+dis[j] > k && i < j)//双指针
               j--;
         ans += j-i;
     }
     return ans;
 }

 //分治，找到树的重心，分为经过重心的点对和不经过重心的点对。
 void solve(int u){
     MAX = n;
     dfs_size(u, );
     dfs_root(u, u, );
     answer += cal(root, );
     vis[root] = ;
     for(int i = head[root]; ~i; i = edge[i].next){
         int v = edge[i].to;
         if(vis[v])continue;
         answer -= cal(v, edge[i].w);
         solve(v);
     }
 }

 int main()
 {
     //freopen("dataIn.txt", "r", stdin);
     while(scanf("%d%d", &n, &k)!=EOF){
         if(!n && !k)break;
         int u, v, w;
         init(n);
         for(int i = ; i < n-; i++){
             scanf("%d%d%d", &u, &v, &w);
             add_edge(u, v, w);
             add_edge(v, u, w);
         }
         solve();
         printf("%d\n", answer);
     }

     return ;
 }