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 \(n\), \(k\). \((n\leq 10000)\) 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
Solution
点分治裸题。简介见点分治简介。注意要去重（见 \(calcPair\) ）。
#include<bits/stdc++.h>
using namespace std;
#define N 100001
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define drp(i, a, b) for (int i = a; i >= b; i--)
#define fech(i, x) for (int i = 0; i < x.size(); i++)
#define pii pair<int, int>
#define INF 0x7fffffff
inline int read() {
int x = 0, flag = 1; char ch = getchar(); while (!isdigit(ch)) { if (!(ch ^ '-')) flag = -1; ch = getchar(); }
while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - '0', ch = getchar(); return x * flag;
}
inline void write(int x) {
if (!x) { putchar('0'); return; } if (x < 0) putchar('-'), x = -x;
char buf[20] = ""; int top = 0; while (x) buf[++top] = x % 10 + '0', x /= 10; while (top) putchar(buf[top--]);
}
int n, K;
struct edgeType { int u, v, w; }eg[N]; int tot;
vector<int> g[N];
bool centr[N];
int Size[N];
int ans;
pii calcCentr(int u, int f, int sz) {
int mx = 0, sum = 1; pii res = pii{ INF, -1 };
fech(i, g[u]) {
edgeType e = eg[g[u][i]]; if (!(e.v ^ f) || centr[e.v]) continue;
res = min(res, calcCentr(e.v, u, sz)); mx = max(mx, Size[e.v]); sum += Size[e.v];
}
mx = max(mx, sz - sum); res = min(res, pii{ mx, u }); return res;
}
int calcSize(int u, int f) {
Size[u] = 1;
fech(i, g[u]) {
edgeType e = eg[g[u][i]]; if (!(e.v ^ f) || centr[e.v]) continue;
calcSize(e.v, u); Size[u] += Size[e.v];
}
return Size[u];
}
void calcDis(int u, int f, int d, vector<int>& ds) {
ds.push_back(d);
fech(i, g[u]) {
edgeType e = eg[g[u][i]]; if (!(e.v ^ f) || centr[e.v]) continue;
calcDis(e.v, u, e.w + d, ds);
}
}
inline int calcPair(vector<int>& ds) {
int res = 0;
sort(ds.begin(), ds.end());
int j = ds.size();
fech(i, ds) {
while (j > 0 && ds[i] + ds[j - 1] > K) j--;
res += j - (j > i ? 1 : 0);
}
return res / 2;
}
void solve(int u) {
calcSize(u, 0);
int s = calcCentr(u, 0, Size[u]).second;
centr[s] = 1;
fech(i, g[u]) if (!centr[eg[g[u][i]].v]) solve(eg[g[u][i]].v);
vector<int> ds; ds.push_back(0);
fech(i, g[u]) {
edgeType e = eg[g[u][i]]; if (centr[e.v]) continue;
vector<int> tds; calcDis(e.v, u, e.w, tds);
ans -= calcPair(tds);
ds.insert(ds.begin(), tds.begin(), tds.end());
}
ans += calcPair(ds);
centr[s] = 0;
}
int main() {
while (scanf("%d%d", &n, &K) == 2 && n && K) {
tot = 0; rep(i, 1, n) g[i].clear();
rep(i, 2, n) {
int u = read(), v = read(), w = read();
eg[++tot] = edgeType{ u, v, w }; g[u].push_back(tot);
eg[++tot] = edgeType{ v, u, w }; g[v].push_back(tot);
}
ans = 0; solve(1); write(ans); puts("");
}
return 0;
}