BZOJ 2152 匆匆可可 | 樹的點分治

點分治板題。

把樹從重心分開，遞歸處理各個子樹，並處理路徑跨重心的情況：計算所有點中距離重心的距離相加為3的倍數的點對數，再減去每個子樹中距離重心距離和加上二倍“重心到該子樹根距離”為3的倍數的點對數。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
#define space putchar(' ')
#define enter putchar('
 
\n')
template <class T>
void read(T &x){
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10
 
 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x){
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}

const int N = 20005, INF = 0x3f3f3f3f;
int n;
int ecnt, adj[N], nxt[2*N], go[2*N], w[2*N];
int fa[N], cnt[3], dis[N], sze[N], son[N], que[N], qr;
bool
 
 vis[N];
ll ans;
void add(int u, int v, int ww){
    go[++ecnt] = v;
    nxt[ecnt] = adj[u];
    adj[u] = ecnt;
    w[ecnt] = ww;
}
int calcG(int u){
    que[qr = 1] = u, fa[u] = 0;
    for(int ql = 1; ql <= qr; ql++){
        u = que[ql], sze[u] = 1, son[u] = 0;
        for(int e = adj[u], v; e; e = nxt[e])
            if(!vis[v = go[e]] && v != fa[u])
                que[++qr] = v, fa[v] = u;
    }
    int ret, mi = INF;
    for(int ql = qr; ql; ql--){
        u = que[ql];
        sze[fa[u]] += sze[u];
        son[fa[u]] = max(son[fa[u]], sze[u]);
        son[u] = max(son[u], qr - sze[u]);
        if(son[u] < mi) ret = u, mi = son[u];
    }
    return ret;
}
ll calc(int u, int l){
    memset(cnt, 0, sizeof(cnt));
    que[qr = 1] = u, dis[u] = l % 3, fa[u] = 0;
    for(int ql = 1; ql <= qr; ql++){
        u = que[ql], cnt[dis[u]]++;
        for(int e = adj[u], v; e; e = nxt[e])
            if(!vis[v = go[e]] && v != fa[u])
                que[++qr] = v, fa[v] = u, dis[v] = (dis[u] + w[e]) % 3;
    }
    return (ll)cnt[0] * cnt[0] + (ll)cnt[1] * cnt[2] * 2;
    //一個是1一個是2的情況，正著反著各算一種，要乘二
}
void solve(int u){
    u = calcG(u);
    vis[u] = 1;
    ans += calc(u, 0);
    for(int e = adj[u]; e; e = nxt[e])
        if(!vis[go[e]]) ans -= calc(go[e], w[e]);
    for(int e = adj[u]; e; e = nxt[e])
        if(!vis[go[e]]) solve(go[e]);
}
ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}
int main(){
    read(n);
    for(int i = 1, u, v, ww; i < n; i++)
        read(u), read(v), read(ww), add(u, v, ww), add(v, u, ww);
    solve(1);
    ll g = gcd(ans, (ll)n * n);
    printf("%lld/%lld\n", ans / g, (ll)n * n / g);
    return 0;
}

BZOJ 2152 聰聰可可 | 樹的點分治