POJ 1741 Tree（树的分治）

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.

题目大意：给一个带边权的树，问有多少对点满足dist(i,j)<=k

思路：可以参考2009年的国家集训队论文《分治算法在树的路径问题中的应用》——漆子超。

代码（188MS）：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 using namespace std;
 
 const int MAXN = ;
 const int MAXE = ;
 const int INF = 0x7fff7fff;
 
 int head[MAXN], size[MAXN], maxSize[MAXN];
 int list[MAXN], cnt;
 bool del[MAXN];
 int to[MAXE], next[MAXE], cost[MAXE];
 int n, k, ecnt;
 
 void init() {
     memset(head, -, sizeof(head));
     memset(del, , sizeof(del));
     ecnt = ;
 }
 
 void add_edge(int u, int v, int c) {
     to[ecnt] = v; cost[ecnt] = c; next[ecnt] = head[u]; head[u] = ecnt++;
     to[ecnt] = u; cost[ecnt] = c; next[ecnt] = head[v]; head[v] = ecnt++;
 }
 
 void dfs1(int u, int f) {
     size[u] = ;
     maxSize[u] = ;
     for(int p = head[u]; ~p; p = next[p]) {
         int &v = to[p];
         if(v == f || del[v]) continue;
         dfs1(v, u);
         size[u] += size[v];
         maxSize[u] = max(maxSize[u], size[v]);
     }
     list[cnt++] = u;
 }
 
 int get_root(int u, int f) {
     cnt = ;
     dfs1(u, f);
     int ret, maxr = INF;
     for(int i = ; i < cnt; ++i) {
         int &x = list[i];
         if(max(maxSize[x], size[u] - size[x]) < maxr) {
             ret = x;
             maxr = max(maxSize[x], size[u] - size[x]);
         }
     }
     return ret;
 }
 
 void dfs2(int u, int f, int dis) {
     list[cnt++] = dis;
     for(int p = head[u]; ~p; p = next[p]) {
         int &v = to[p];
         if(v == f || del[v]) continue;
         dfs2(v, u, dis + cost[p]);
     }
 }
 
 int calc(int a, int b) {
     int j = b - , ret = ;
     for(int i = a; i < b; ++i) {
         while(list[i] + list[j] > k && i < j) --j;
         ret += j - i;
         if(j == i) break;
     }
     return ret;
 }
 
 int ans = ;
 
 void work(int u, int f) {
     int root = get_root(u, f);
     del[root] = true;
     int last = ; cnt = ;
     for(int p = head[root]; ~p; p = next[p]) {
         int &v = to[p];
         if(del[v]) continue;
         dfs2(v, root, cost[p]);
         sort(list + last, list + cnt);
         ans -= calc(last, cnt);
         last = cnt;
     }
     list[cnt++] = ;
     sort(list, list + cnt);
     ans += calc(, cnt);
     for(int p = head[root]; ~p; p = next[p]) {
         int &v = to[p];
         if(del[v]) continue;
         work(v, root);
     }
 }
 
 int main() {
     while(scanf("%d%d", &n, &k) != EOF) {
         if(n ==  && k == ) break;
         init();
         for(int i = ; i < n; ++i) {
             int u, v, c;
             scanf("%d%d%d", &u, &v, &c);
             add_edge(u, v, c);
         }
         ans = ;
         work(, );
         printf("%d\n", ans);
     }
 }