题解
距离省选只有一周了我居然才开始数据结构康复计划
这题很简单，就是点分树，然后二分一个值，我们计算有多少条路径大于这个值
对于一个点分树上的重心，我们可以通过双指针的方法求出它子树里的路径任意搭配大于这个值的方案
然后同一个子树里重复计算的删掉
再计算和自己祖先之间的路径
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 50005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 +c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
struct node {
int to,next,val;
}E[MAXN * 2];
int N,sumE,head[MAXN],rt;
int64 K;
int dep[MAXN],fa[MAXN][20],d[MAXN],zz[MAXN];
bool vis[MAXN];
vector<int> poi;
vector<int> dis[2][MAXN];
int tot;
void add(int u,int v,int c) {
E[++sumE].to = v;
E[sumE].next = head[u];
E[sumE].val = c;
head[u] = sumE;
}
void dfs(int u) {
zz[u] = zz[fa[u][0]] + 1;
for(int i = 1 ; i <= 18 ; ++i) fa[u][i] = fa[fa[u][i - 1]][i - 1];
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != fa[u][0]) {
fa[v][0] = u;
dep[v] = dep[u] + E[i].val;
dfs(v);
}
}
}
int que[MAXN],ql,qr;
int Calc_G(int st) {
static int f[MAXN],siz[MAXN],son[MAXN];
ql = 1;qr = 0;
que[++qr] = st;f[st] = 0;siz[st] = 1;son[st] = 0;
while(ql <= qr) {
int u = que[ql++];
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != f[u] && !vis[v]) {
siz[v] = 1;son[v] = 0;
f[v] = u;que[++qr] = v;
}
}
}
int res = que[qr];
for(int i = qr ; i >= 1 ; --i) {
int u = que[i];
if(f[u]) {
siz[f[u]] += siz[u];
son[f[u]] = max(son[f[u]],siz[u]);
}
son[u] = max(son[u],qr - siz[u]);
if(son[u] < son[res]) res = u;
}
return res;
}
void dfs_for_dis(int u,int f) {
poi.pb(u);
for(int i = head[u] ; i ; i = E[i].next) {
int v = E[i].to;
if(v != f && !vis[v]) {
d[v] = d[u] + E[i].val;
dfs_for_dis(v,u);
}
}
}
void dfs_divide(int u) {
int G = Calc_G(u);
vis[G] = 1;
for(int i = head[G] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v]) {
poi.clear();
d[v] = E[i].val;
dfs_for_dis(v,G);
++tot;
for(int j = 0 ; j < poi.size() ; ++j) {
dis[1][tot].pb(d[poi[j]]);
dis[0][G].pb(d[poi[j]]);
}
}
}
for(int i = head[G] ; i ; i = E[i].next) {
int v = E[i].to;
if(!vis[v]) dfs_divide(v);
}
}
int64 check(int mid) {
int64 res = 0;
for(int i = 1 ; i <= N ; ++i) {
int l = 0,r = dis[0][i].size() - 1;
while(r >= 0) {
while(l < dis[0][i].size() && dis[0][i][r] + dis[0][i][l] < mid) ++l;
if(dis[0][i][r] >= mid) res += 2;
res += dis[0][i].size() - l;
--r;
}
l = 0,r = dis[1][i].size() - 1;
while(r >= 0) {
while(l < dis[1][i].size() && dis[1][i][r] + dis[1][i][l] < mid) ++l;
res -= dis[1][i].size() - l;
--r;
}
}
res /= 2;
for(int i = 1 ; i <= N ; ++i) {
int t = dep[i] - mid;
int u = i;
for(int j = 18 ; j >= 0 ; --j) {
if(fa[u][j] && dep[fa[u][j]] > t) {
u = fa[u][j];
}
}
res -= zz[u] - zz[rt];
}
return res;
}
void Init() {
read(N);read(rt);read(K);
memset(head,0,sizeof(head));
memset(vis,0,sizeof(vis));
sumE = 0;tot = 0;
for(int i = 1 ; i <= N ; ++i) {dis[0][i].clear();dis[1][i].clear();}
int u,v,c;
for(int i = 1 ; i < N ; ++i) {
read(u);read(v);read(c);
add(u,v,c);add(v,u,c);
}
dep[rt] = 0;
fa[rt][0] = 0;
dfs(rt);
dfs_divide(rt);
for(int i = 1 ; i <= N ; ++i) {
sort(dis[0][i].begin(),dis[0][i].end());
sort(dis[1][i].begin(),dis[1][i].end());
}
}
void Solve() {
Init();
int L = 0,R = 500000001;
while(L < R) {
int mid = (L + R + 1) >> 1;
if(check(mid) >= K) L = mid;
else R = mid - 1;
}
if(R == 0) {puts("NO");return;}
out(L);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
int T;
read(T);
while(T--) Solve();
}