POJ 1741 Tree(点分治)

点分治...模板题吧..只要理解了,实现并不难,随便搞搞...

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<vector>

 

#define rep(i, n) for(int i = 0; i < n; ++i)

#define REP(x) for(edge* e = head[x]; e; e = e->next)

#define clr(x, c) memset(x, c, sizeof(x))

 

using namespace std;

 

const int maxn = 10000 + 5;

const int inf = 0x7fffffff;

 

int n, K;

 

struct edge {

int to, dist;

edge* next;

};

 

edge* pt;

edge* head[maxn];

edge EDGE[maxn << 1];

 

inline void init() {

pt = EDGE;

clr(head, 0);

}

 

inline void add(int u, int v, int d) {

pt->to = v;

pt->dist = d;

pt->next = head[u];

head[u] = pt++;

}

 

inline void add_edge(int u, int v, int d) {

add(u, v, d);

add(v, u, d);

}

 

int size[maxn];

bool vis[maxn];

 

int dp(int x, int fa) {

size[x] = 1;

REP(x) if(!vis[e->to] && e->to != fa)

size[x] += dp(e->to, x);

return size[x];

}

 

int root, Min;

int node_num;

 

void dfs(int x, int fa) {

int Max = node_num - size[x];

REP(x) if(fa != e->to && !vis[e->to]) {

Max = max(Max, size[e->to]);

dfs(e->to, x);

}

if(Max < Min)

   root = x, Min = Max;

}

 

void Dfs(int x) {

Min = inf;

node_num = dp(x, -1);

dfs(x, -1);

}

 

vector<int> d;

 

void DFS(int x, int fa, int dist) {

d.push_back(dist);

REP(x) if(!vis[e->to] && fa != e->to)

DFS(e->to, x, dist + e->dist);

}

 

int Count(int x, int dist) {

d.clear();

DFS(x, -1, dist);

sort(d.begin(), d.end());

int ans = 0, l = 0, r = d.size() - 1;

while(l < r) {

while(d[l] + d[r] > K && l < r) r--;

ans += r - l;

l++;

}

return ans;

}

 

int ANS;

 

void COUNT(int x) {

Dfs(x);

vis[root] = 1;

ANS += Count(root, 0);

REP(root) if(!vis[e->to])

   ANS -= Count(e->to, e->dist), COUNT(e->to);

}

 

inline void COUNT_init() {

ANS = 0;

clr(vis, 0);

}

 

void work() {

COUNT_init();

COUNT(0);

cout << ANS << " ";

}

 

void Read() {

int u, v, d;

rep(i, n - 1)

   scanf("%d%d%d", &u, &v, &d), add_edge(u - 1, v - 1, d);

}

 

int main() {

// freopen("test.in", "r", stdin);

while(cin >> n >> K) {

if(!n && !K) break;

init();

Read();

work();

}

return 0;

}

  

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Tree

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12802		Accepted: 4079

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

LouTiancheng@POJ