LuoguP3806 【模板】点分治1 (点分治)

给定一棵有n个点的树，询问树上距离为k的点对是否存在。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long
#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("

----------

")

#else

#define D_e_Line ;

#endif

struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std; 
const int N = 10007;
const int MAXX = 10000000;

struct Edge{
	int nxt,pre,w;
}e[N<<1];
int cntEdge,head[N];
inline void add(int u,int v,int w){
	e[++cntEdge] = (Edge){head[u], v, w}, head[u] = cntEdge;
}

int minn, root, treeSize;

int vis[N],siz[N];
inline void GetRoot(int u,int fa){
	siz[u] = 1;
	int sum = 0;
	for(register int i = head[u]; i; i = e[i].nxt){
		int v = e[i].pre;
		if(vis[v] || v == fa) continue;
		GetRoot(v, u);
		sum = Max(sum, siz[v]);
		siz[u] += siz[v];
	}
	sum = Max(sum, treeSize - siz[u]);
	if(sum < minn){
		minn = sum;
		root = u;
	}
}

int dis[N];

int tot;
inline void GetDistance(int u,int fa,int w){
	dis[++tot] = w;
	for(register int i = head[u]; i; i = e[i].nxt){
		int v = e[i].pre;
		if(vis[v] || v == fa) continue;
		GetDistance(v, u, w + e[i].w);
	}
}

int num[MAXX];
inline void Solve(int u,int w,int flag){ // w is the influence of the beginning node
	tot = 0;
	GetDistance(u, u, 0);
	R(i,1,tot){
		R(j,1,tot){
			if(flag && dis[i] + dis[j] <= MAXX)
				++ num[dis[i] + dis[j]];
			else
				if(dis[i] + dis[j] +w <= MAXX)
					-- num[dis[i] + dis[j] + w];
		}
	}
}

inline void DFS(int u){
	Solve(u, 0, 1);
	vis[u] = true;
	for(register int i = head[u]; i; i = e[i].nxt){
		int v = e[i].pre;
		if(vis[v]) continue;
		Solve(v, (e[i].w << 1), 0); // QAQ //  T^T /
		treeSize = siz[v];
		minn = 0x7fffffff;
		GetRoot(v, u);
		DFS(root);
	}
}

int main(){
	int n,m;
	io >> n >> m;
	R(i,2,n){
		int u,v,w;
		io >> u >> v >> w;
		add(u, v, w);
		add(v, u, w);
	}
	
	treeSize = n; // because I forgot this sentence, the whole morning...
	minn = 0x7fffffff;
	GetRoot(1, 1);
	DFS(root);
	
	while(m--){
		int x;
		io >> x;
		if(num[x])
			printf("AYE
");
		else
			printf("NAY
");
	}
	
	return 0;
}

a faster way in some cases

inline int Solve(int u,int w){
	tot = 0;
	GetDistance(u, u, 0);
	sort(dis + 1, dis + tot + 1);
	int l = 0, r = tot;
	while(l < r){
		while(l < r && dis[l] + dis[r] > MAXX) -- r;
		ans += r - l;
		++ l;
	}
	return ans;
}