P4178 Tree

(color{#0066ff}{题目描述})

给你一棵TREE,以及这棵树上边的距离.问有多少对点它们两者间的距离小于等于K

(color{#0066ff}{输入格式})

N（n<=40000） 接下来n-1行边描述管道，按照题目中写的输入 接下来是k

(color{#0066ff}{输出格式})

一行，有多少对点之间的距离小于等于k

(color{#0066ff}{输入样例})

7
1 6 13 
6 3 9 
3 5 7 
4 1 3 
2 4 20 
4 7 2 
10

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5

(color{#0066ff}{数据范围与提示})

none

(color{#0066ff}{题解})

点分治

拆掉重心，对子树操作

维护一个权值树状数组

搜完一棵子树，把所有的dis放到一个数组里

for在树状数组上统计答案（注意要+1，那是当前点与重心的点对贡献）

之后把当前子树加入树状数组，最后清空

#include<bits/stdc++.h>
using namespace std;
#define LL long long
LL in() {
	char ch; int x = 0, f = 1;
	while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
	for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
	return x * f;
}
struct node {
	int to, dis;
	node *nxt;
	node(int to = 0, int dis = 0, node *nxt = NULL): to(to), dis(dis), nxt(nxt) {}
	void *operator new (size_t) {
		static node *S = NULL, *T = NULL;
		return (S == T) && (T = (S = new node[1024]) + 1024), S++;
	}
};
const int maxn = 40505;
int siz[maxn], f[maxn], tmp[maxn], ls[maxn];
node *head[maxn];
bool vis[maxn];
int n, k, num, cnt, sum, root;
int ans;
struct BIT {
protected:
	int st[maxn];
	int low(int x) { return x & (-x); }
public:
	void add(int pos, int x) { while(pos <= k) st[pos] += x, pos += low(pos); }
	int query(int pos) { int re = 0; while(pos) re += st[pos], pos -= low(pos); return re; }
}b;
void add(int from, int to, int dis) {
	head[from] = new node(to, dis, head[from]);
}
void getdis(int x, int fa, int dis) {
	ls[++cnt] = dis;
	for(node *i = head[x]; i; i = i->nxt) 
		if(i->to != fa && !vis[i->to])
			getdis(i->to, x, dis + i->dis);
}
void getroot(int x, int fa) {
	f[x] = 0, siz[x] = 1;
	for(node *i = head[x]; i; i = i->nxt) {
		if(vis[i->to] || i->to == fa) continue;
		getroot(i->to, x);
		siz[x] += siz[i->to];
		f[x] = std::max(f[x], siz[i->to]);
	}
	f[x] = std::max(f[x], sum - siz[x]);
	if(f[x] < f[root]) root = x;
}
void calc(int x) {
	num = 0;
	for(node *i = head[x]; i; i = i->nxt) {
		if(vis[i->to]) continue;
		cnt = 0;
		getdis(i->to, 0, i->dis);
		for(int j = 1; j <= cnt; j++) if(ls[j] <= k) ans += b.query(k - ls[j]) + 1;
		for(int j = 1; j <= cnt; j++) if(ls[j] <= k) b.add(ls[j], 1), tmp[++num] = ls[j];
	}
	for(int i = 1; i <= num; i++) b.add(tmp[i], -1);
}

void work(int x) {
	vis[x] = true;
	calc(x);
	for(node *i = head[x]; i; i = i->nxt) {
		if(vis[i->to]) continue;
		root = 0, sum = siz[i->to];
		getroot(i->to, 0);
		work(root);
	}
}
int main() {
	n = in();
	int x, y, z;
	for(int i = 1; i < n; i++) {
		x = in(), y = in(), z = in();
		add(x, y, z), add(y, x, z);
	}
	k = in();
	f[0] = sum = n;
	getroot(1, 0); 
	work(root);
	printf("%d", ans);
	return 0;
}