POJ 1741 Tree——点分治

题意：给定一棵树，求合法点对的个数，合法点对定义为：两点之间的距离不超过k

思路：

1求树的重心

2以树的重心为根，求此次划分中其余的点到根节点的距离

3计算此次划分的合法点对

4减去在同一子树中的点对（因为这些点对在后面的划分中会重复统计），然后断开根节点，对子树重复1的过程

复杂度为n（logn）^2

点分治的更多介绍参考09年国家集训队论文

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e4+5;
const int INF = 0x3f3f3f3f;
bool vis[maxn];
int n, k, ans, tot, head[maxn], sz[maxn], dp[maxn], root, dis[maxn], cnt, SIZE;
struct Edge { int to, val, next; }edge[maxn<<1];
void init() {
    ans = tot = 0;
    memset(head, -1, sizeof(head));
    memset(vis, false, sizeof(vis));
}
void addedge(int u, int v, int val) {
    edge[tot].to = v; edge[tot].val = val; edge[tot].next = head[u];
    head[u] = tot++;
}
void getroot(int u, int pre) {
    sz[u] = 1, dp[u] = 0;
    for (int i = head[u]; ~i; i = edge[i].next) {
        int v = edge[i].to;
        if (vis[v] || v == pre) continue;
        getroot(v, u);
        sz[u] += sz[v];
        dp[u] = max(dp[u], sz[v]);
    }
    dp[u] = max(dp[u], SIZE-sz[u]);
    if (dp[u] < dp[root]) root = u;
}
void getdis(int u, int pre, int d) {
    dis[++cnt] = d;
    sz[u] = 1;
    for (int i = head[u]; ~i; i = edge[i].next) {
        int v = edge[i].to, val = edge[i].val;
        if (vis[v] || v == pre) continue;
        getdis(v, u, d + val);
        sz[u] += sz[v];
    }
}
int getans(int u, int d) {
    cnt = 0;
    int sum = 0;
    getdis(u, -1, d);
    sort(dis+1, dis+1+cnt);
    int i = 1, j = cnt;
    while (i < j) {
        while (dis[i]+dis[j] > k && i < j) j--;
        sum += j - i; i++;
    }
    return sum;
}
void solve(int u) {
    dp[0] = INF, root = 0;
    getroot(u, -1);
    ans += getans(root, 0);
    vis[root] = true;
    for (int i = head[root]; ~i; i = edge[i].next) {
        int v = edge[i].to, val = edge[i].val;
        if (vis[v]) continue;
        ans -= getans(v, val);
        SIZE = sz[v];
        solve(v);
    }
}

int main() {
    while (~scanf("%d %d", &n, &k) && (n+k)) {
        int u, v, val;
        init();
        for (int i = 1; i <= n-1; i++) {
            scanf("%d %d %d", &u, &v, &val);
            addedge(u, v, val); addedge(v, u, val);
        }
        SIZE = n;
        solve(1);
        printf("%d\n", ans);
    }
    return 0;
}