POJ1741Tree(树上点分治)

Tree Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 13091 Accepted: 4205 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. Sample Input 5 4 1 2 3 1 3 1 1 4 2 3 5 1 0 0 Sample Output 8 Source [email protected]

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第一道点分治TAT

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e4 + 10;
const int inf = 1e8;
int siz[maxn],f[maxn],root,tot,k;
int head[maxn],e,dist[maxn],d[maxn];
bool vis[maxn];
struct Edge
{
	int to,w,next;
}G[maxn<<1];
void AddEdge(int u,int v,int w)
{
	G[++e] = (Edge) {v,w,head[u]};head[u] = e;
	G[++e] = (Edge) {u,w,head[v]};head[v] = e;
}
void getroot(int u,int fa)
{
	siz[u] = 1;f[u] = 0;
	for(int eid = head[u]; eid ; eid = G[eid].next) {
		int v = G[eid].to;
		if(v==fa||vis[v])continue;
		getroot(v,u);
		siz[u] += siz[v];
		f[u] = max(f[u],siz[v]);
	}
	f[u] = max(f[u],tot - siz[u]);
	if(f[u] < f[root]) root = u;
}
void getdist(int u,int fa)
{
	dist[++dist[0]] = d[u];
	for(int eid = head[u]; eid ; eid = G[eid].next) {
		int v = G[eid].to;
		if(v==fa||vis[v])continue;
		d[v] = d[u] + G[eid].w;
		getdist(v,u);
	}
}
int cacu(int u,int now)
{
	dist[0] = 0;
	d[u] = now;
	getdist(u,0);
	sort(dist+1,dist+1+dist[0]);
	int l = 1,r = dist[0];
	int num = 0;
	while(l < r) {
		if(dist[l] + dist[r] <= k) {
			num += r - l;
			l++;
		}else r--;
	}
	return num;
}
int solve(int u)
{
	int ans = cacu(u,0);vis[u] = 1;
	for(int eid = head[u]; eid ; eid = G[eid].next) {
		int v = G[eid].to;
		if(vis[v]) continue;
		ans -= cacu(v,G[eid].w);
		root = 0;
		tot = siz[v];
		getroot(v,u);
		ans += solve(root);
	}
	return ans;
}
int main(int argc, char const *argv[])
{
	int n;
	while(scanf("%d%d",&n,&k)==2) {
		if(!n)return 0;
		for(int i = 1; i <= n; i++)head[i] = 0;
		for(int i = 1; i <= n; i++)vis[i] = 0;
		e = 0;
		for(int i = 1; i < n; i++) {
			int u,v,w;scanf("%d%d%d",&u,&v,&w);
			AddEdge(u,v,w);
		}
		f[root = 0] = inf;
		tot = n;
		getroot(1,-1);
		int ans = solve(root);
		printf("%d\n", ans);
	}
	return 0;
}