51Nod-1811-联通分量计数

来源：互联网发布：linux depth 编辑：程序博客网时间：2024/06/05 16:09

ACM模版

描述

题解

感觉这个题好难啊，虽然知道是要求每条边的贡献，但是完全不知道具体怎么搞，花了 5 盾看了题解……

虽说是思路上理解了，但是后边的 启发式合并 + 数据结构来维护子树 还是一脸懵逼，于是一狠心，又花了 60 盾看了大牛们的代码……好吧，我必须承认，没有看懂，大体上算是理解了一丢丢，但是具体的还是十分头疼……感觉好麻烦的说，树归部分倒是十分容易看懂，但是还是无法顿悟 启发式合并 + 数据结构来维护子树 部分的高超技艺……

Mark 一下，如果有大神有时间写更加详细的题解，烦请留言区留地址……就此谢过~~~

是时候看看 启发式合并 是啥东西了~~~害怕.jpg

代码

#include <cstdio>#include <iostream>#define ll long longusing namespace std;const int MAXN = 2e5 + 5;const int MAXM = MAXN << 4;int n;int cnt = 0, pos = 0;ll ans = 0, tmp;int root[MAXN];int l1[MAXM], r1[MAXM];int ls[MAXM], rs[MAXM];int lb[MAXM], rb[MAXM];int nt[MAXN], head[MAXN], v[MAXN];ll s[MAXM];template <class T>inline void scan_d(T &ret){    char c;    ret = 0;    while ((c = getchar()) < '0' || c > '9');    while (c >= '0' && c <= '9')    {        ret = ret * 10 + (c - '0'), c = getchar();    }}ll get_n(int n){    return (ll)(n + 1) * n / 2;}void add(int x, int y){    nt[++pos] = head[x];    head[x] = pos;    v[pos] = y;}void ins(int &x, int l, int r, int a){    if (!x)    {        x = ++cnt;    }    if (l == r)    {        ls[x] = rs[x] = 1;        lb[x] = rb[x] = 0;        s[x] = 1;        return ;    }    int m = (l + r) >> 1;    if (a <= m)    {        ins(l1[x], l, m, a);    }    else    {        ins(r1[x], m + 1, r, a);    }    if (!l1[x])    {        l1[x] = ++cnt;        lb[cnt] = rb[cnt] = m - l + 1;        s[cnt] = get_n(m - l + 1);    }    if (!r1[x])    {        r1[x] = ++cnt;        lb[cnt] = rb[cnt] = r - m;        s[cnt] = get_n(r - m);    }    int k1 = l1[x], k2 = r1[x];    s[x] = s[k1] + s[k2];    if (ls[k2] && rs[k1])    {        s[x] -= get_n(rs[k1]) + get_n(ls[k2]);        s[x] += get_n(rs[k1] + ls[k2]);    }    if (lb[k2] && rb[k1])    {        s[x] -= get_n(rb[k1]) + get_n(lb[k2]);        s[x] += get_n(rb[k1] + lb[k2]);    }    ls[x] = ls[k1];    rs[x] = rs[k2];    lb[x] = lb[k1];    rb[x] = rb[k2];    if (ls[k1] == m - l + 1)    {        ls[x] += ls[k2];    }    if (lb[k1] == m - l + 1)    {        lb[x] += lb[k2];    }    if (rs[k2] == r - m)    {        rs[x] += rs[k1];    }    if (rb[k2] == r - m)    {        rb[x] += rb[k1];    }}int merge(int x, int y, int l, int r){    if (!x || !y)    {        return x + y;    }    if (lb[x] == r - l + 1)    {        return y;    }    if (lb[y] == r - l + 1)    {        return x;    }    int m = (l + r) >> 1;    l1[x] = merge(l1[x], l1[y], l, m);    r1[x] = merge(r1[x], r1[y], m + 1, r);    if (!l1[x])    {        l1[x] = ++cnt;        lb[cnt] = rb[cnt] = m - l + 1;        s[cnt] = get_n(m - l + 1);    }    if (!r1[x])    {        r1[x] = ++cnt;        lb[cnt] = rb[cnt] = r - m;        s[cnt] = get_n(r - m);    }    int k1 = l1[x], k2 = r1[x];    s[x] = s[k1] + s[k2];    if (ls[k2] && rs[k1])    {        s[x] -= get_n(rs[k1]) + get_n(ls[k2]);        s[x] += get_n(rs[k1] + ls[k2]);    }    if (lb[k2] && rb[k1])    {        s[x] -= get_n(rb[k1]) + get_n(lb[k2]);        s[x] += get_n(rb[k1] + lb[k2]);    }    ls[x] = ls[k1];    rs[x] = rs[k2];    lb[x] = lb[k1];    rb[x] = rb[k2];    if (ls[k1] == m - l + 1)    {        ls[x] += ls[k2];    }    if (lb[k1] == m - l + 1)    {        lb[x] += lb[k2];    }    if (rs[k2] == r - m)    {        rs[x] += rs[k1];    }    if (rb[k2] == r - m)    {        rb[x] += rb[k1];    }    return x;}void dfs(int rt, int pre){    for (int i = head[rt]; i; i = nt[i])    {        int v_ = v[i];        if (v_ != pre)        {            dfs(v_, rt);            root[rt] = merge(root[rt], root[v_], 1, n);        }    }    ins(root[rt], 1, n, rt);    ans += tmp - s[root[rt]];}int main(){    scan_d(n);    tmp = get_n(n);    int x, y;    for (int i = 1; i < n; i++)    {        scan_d(x), scan_d(y);        add(x, y), add(y, x);    }    dfs(1, 0);    printf("%lld\n", ans);    return 0;}

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