HDU - 3534 Tree

题意：

给定一棵有 $n$ 个结点的树，求两点间最长距离，以及最长距离的点对个数。 $(n\leq~?)$

链接：

https://vjudge.net/problem/HDU-3534

解题思路：

即求直径，以及直径的个数。直径可以由两次 $dfs$ 得到，距离为 $d$ 的点对个数，点分可解。更简单的做法是树形 $dp$，自下而上维护最长链及最长链个数，相应更新。

参考代码：

树形 $dp$：

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define sz(a) ((int)a.size())
#define pb push_back
#define lson (rt << 1)
#define rson (rt << 1 | 1)
#define gmid (l + r >> 1)
const int maxn = 1e6 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;

vector<pii> G[maxn];
ll mx[maxn], sum[maxn];
int n; ll ans1, ans2;

void dfs(int u, int f){

    mx[u] = 0, sum[u] = 1;
    for(auto &e : G[u]){

        int v = e.second, w = e.first;
        if(v == f) continue;
        dfs(v, u);
        if(mx[v] + w + mx[u] > ans1) ans1 = mx[v] + w + mx[u], ans2 = sum[u] * sum[v];
        else if(mx[v] + w + mx[u] == ans1) ans2 += sum[u] * sum[v];
        if(mx[v] + w > mx[u]) mx[u] = mx[v] + w, sum[u] = sum[v];
        else if(mx[v] + w == mx[u]) sum[u] += sum[v];
    }
}

int main(){

    ios::sync_with_stdio(0); cin.tie(0);
    while(cin >> n){

        ans1 = ans2 = 0;
        for(int i = 1; i <= n; ++i){

            G[i].clear();
        }
        for(int i = 1; i < n; ++i){

            int u, v, w; cin >> u >> v >> w;
            G[u].pb({w, v}), G[v].pb({w, u});
        }
        dfs(1, 0);
        cout << ans1 << " " << ans2 << endl;
    }
    return 0;
}

 
点分治：

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define sz(a) ((int)a.size())
#define pb push_back
#define lson (rt << 1)
#define rson (rt << 1 | 1)
#define gmid (l + r >> 1)
const int maxn = 1e6 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;

vector<pii> G[maxn];
int siz[maxn], vis[maxn];
ll dis[maxn], ans1, ans2;
int n, tn, rmn, rt, tot;

void getRt(int u, int f){

    int mx = 0; siz[u] = 1;
    for(auto &e : G[u]){

        int v = e.second, w = e.first;
        if(v == f || vis[v]) continue;
        getRt(v, u);
        siz[u] += siz[v];
        mx = max(mx, siz[v]);
    }
    mx = max(mx, tn - siz[u]);
    if(mx < rmn) rmn = mx, rt = u;
}

void dfs(int u, int f, ll d){

    dis[++tot] = d;
    for(auto &e : G[u]){

        int v = e.second, w = e.first;
        if(v == f || vis[v]) continue;
        dfs(v, u, d + w);
    }
}

void cal(int u, int d, int flg){

    tot = 0;
    dfs(u, 0, d);
    sort(dis + 1, dis + 1 + tot);
    for(int i = 1; i <= tot; ++i){

        ll tmp = ans1 - dis[i];
        int p1 = lower_bound(dis + 1, dis + 1 + tot, tmp) - dis;
        int p2 = lower_bound(dis + 1, dis + 1 + tot, tmp + 1) - dis;
        ans2 += flg * (p2 - p1);
    }
}

void dfz(int u){

    vis[u] = 1;
    cal(u, 0, 1);
    for(auto &e : G[u]){

        int v = e.second, w = e.first;
        if(vis[v]) continue;
        cal(v, w, -1);
        tn = siz[v], rmn = inf, getRt(v, u);
        dfz(rt);
    }
    vis[u] = 0;
}

void dfss(int u, int f){

    for(auto &e : G[u]){

        int v = e.second, w = e.first;
        if(v == f) continue;
        dis[v] = dis[u] + w;
        dfss(v, u);
    }
}

ll getLen(){

    dis[1] = 0, dfss(1, 0);
    int u = max_element(dis + 1, dis + 1 + n) - dis;
    dis[u] = 0, dfss(u, 0);
    ll d = *max_element(dis + 1, dis + 1 + n);
    return d;
}

int main(){

    ios::sync_with_stdio(0); cin.tie(0);
    while(cin >> n){

        ans1 = ans2 = 0;
        for(int i = 1; i <= n; ++i){
            
            G[i].clear();
        }
        for(int i = 1; i < n; ++i){

            int u, v, w; cin >> u >> v >> w;
            G[u].pb({w, v}), G[v].pb({w, u});
        }
        ans1 = getLen();
        tn = n, rmn = inf, getRt(1, 0);
        dfz(rt);
        ans2 >>= 1;
        cout << ans1 << " " << ans2 << endl;
    }
    return 0;
}