51Nod-1681-公共祖先

描述

题解

代码

#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

const int MAXN = 1e5 + 10;
const int MAXM = 2e6 + 10;

int n, tot, sz;
int LS[MAXN], RS[MAXN]; //  LS[i]: L 树的 i 结点的 dfs 序的起点 RS[i]: 同理
int LE[MAXN], RE[MAXN]; //  LE[i]: L 树的 i 结点的 dfs 序的终点 RE[i]: 同理
int tmp[MAXN];          //  tmp[i]: R 树中 dfs 序为 i 的结点 u 在 L 树 dfs 序中对应的位置
int pos[MAXN];          //  pos[i]: L 树中 i 结点的 dfs 序的位置
int vis[MAXN];
int head[MAXN];
int root[MAXN];

struct node
{
int l, r, c;
} tree[MAXM];

struct Edge
{
int to, nt;
} e[MAXN];

void init()
{
tot = 0;
memset(head, -1, sizeof(head));
memset(vis, 0, sizeof(vis));
}

void addedge(int u, int v)
{
e[tot].to = v;
e[tot].nt = head[u];
head[u] = tot++;
}

void dfs(int u, int pre, int flag)
{
if (flag)
{
RS[u] = ++sz;
tmp[sz] = pos[u];
}
else
{
LS[u] = ++sz;
pos[u] = sz;
}

int v;
for (int i = head[u]; i != -1; i = e[i].nt)
{
v = e[i].to;
if (v != pre)
{
dfs(v, u, flag);
}
}
if (flag)
{
RE[u] = sz;
}
else
{
LE[u] = sz;
}
}

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();
}
}

void update(int &rt, int l, int r, int num)
{
tree[++sz] = tree[rt];
rt = sz;
tree[rt].c++;

if (l == r)
{
return ;
}
int mid = (l + r) >> 1;
if (num <= mid)
{
update(tree[rt].l, l, mid, num);
}
else
{
update(tree[rt].r, mid + 1, r, num);
}
}

int query(int x, int y, int L, int R, int l, int r)
{
if (L <= l && R >= r)
{
return tree[y].c - tree[x].c;
}
int m = (l + r) >> 1;
int ret = 0;
if (L <= m)
{
ret += query(tree[x].l, tree[y].l, L, R, l, m);
}
if (R > m)
{
ret += query(tree[x].r, tree[y].r, L, R, m + 1, r);
}
return ret;
}

int main()
{
scan_d(n);

int u, v, r = 0;

init();
for (int i = 1; i < n; i++)
{
scan_d(u), scan_d(v);
addedge(u, v);
vis[v] = 1;
}

sz = 0;
for (int i = 1; i <= n; i++)
{
if (vis[i] == 0)
{
r = i;
break;
}
}
dfs(r, 0, 0);

init();
for (int i = 1; i < n; i++)
{
scan_d(u), scan_d(v);
addedge(u, v);
vis[v] = 1;
}

sz = 0;
for (int i = 1; i <= n; i++)
{
if (vis[i] == 0)
{
r = i;
break;
}
}
dfs(r, 0, 1);

root[0] = 0, sz = 0;
for (int i = 1; i <= n; i++)
{
root[i] = root[i - 1];
update(root[i], 1, n, tmp[i]);
}

long long ans = 0;
for (int i = 1; i <= n; i++)
{
int x = query(root[RS[i] - 1], root[RE[i]], LS[i], LE[i], 1, n);
ans += 1LL * (x - 1) * (x - 2) / 2;
}

printf("%lld\n",ans);

return 0;
}

参考