51Nod-1811-联通分量计数

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 long

using 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;
}