【洛谷4219】[BJOI2014]大融合（线段树分治）
题目：
分析：
很明显，查询的是删掉某条边后两端点所在连通块大小的乘积。
有加边和删边，想到LCT。但是我不会用LCT查连通块大小啊。果断弃了
有加边和删边，还跟连通性有关，于是开始yy线段树分治做法（不知道线段树分治？推荐一个伪模板题BZOJ4025二分图事实上这个链接是指向我的博客的）。把每次操作（加边或查询）看做一个时刻，一条边存在的区间就是它加入后没有被查询的时间区间的并。于是用可撤销并查集维护一下连通块大小即可。
代码：
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <map>
#include <cassert>
#undef i
#undef j
#undef k
#undef min
#undef max
#undef true
#undef false
#undef swap
#undef sort
#undef if
#undef for
#undef while
#undef printf
#undef scanf
#undef putchar
#undef getchar
#define _ 0
using namespace std;
namespace zyt
{
template<typename T>
inline bool read(T &x)
{
char c;
bool f = false;
x = 0;
do
c = getchar();
while (c != EOF && c != '-' && !isdigit(c));
if (c == EOF)
return false;
if (c == '-')
f = true, c = getchar();
do
x = x * 10 + c - '0', c = getchar();
while (isdigit(c));
if (f)
x = -x;
return true;
}
inline bool read(char &c)
{
do
c = getchar();
while (c != EOF && !isgraph(c));
return c != EOF;
}
template<typename T>
inline void write(T x)
{
static char buf[20];
char *pos = buf;
if (x < 0)
putchar('-'), x = -x;
do
*pos++ = x % 10 + '0';
while (x /= 10);
while (pos > buf)
putchar(*--pos);
}
typedef long long ll;
const int N = 1e5 + 10, B = 17, QUERY = 0, ADD = 1;
int n, q;
ll ans[N];
struct node
{
bool type;
int x, y;
}arr[N];
namespace UFS
{
int fa[N], rk[N], size[N], top;
struct node
{
int x, y, fax, rky, sizey;
}stack[N];
void init(const int n)
{
for (int i = 1; i <= n; i++)
fa[i] = i, rk[i] = size[i] = 1;
}
int f(const int u)
{
return fa[u] == u ? u : f(fa[u]);
}
bool merge(const int u, const int v)
{
int x = f(u), y = f(v);
if (x == y)
return false;
if (rk[x] > rk[y])
swap(x, y);
stack[top++] = (node){x, y, fa[x], rk[y], size[y]};
fa[x] = y, size[y] += size[x];
if (rk[x] == rk[y])
++rk[y];
return true;
}
int query(const int u)
{
assert(f(u) < N);
return size[f(u)];
}
void undo(const int bck)
{
while (top > bck)
{
--top;
int x = stack[top].x, y = stack[top].y;
assert(x < N && y < N);
fa[x] = stack[top].fax;
rk[y] = stack[top].rky;
size[y] = stack[top].sizey;
}
}
}
namespace Segment_Tree
{
struct edge
{
int x, y, next;
}e[N * (B + 1)];
int head[1 << (B + 1) | 11], ecnt;
inline void init()
{
memset(head, -1, sizeof(head));
}
inline void add(const int a, const int b, const int c)
{
e[ecnt] = (edge){b, c, head[a]}, head[a] = ecnt++;
}
inline void insert(const int rot, const int lt, const int rt, const int ls, const int rs, const int x, const int y)
{
if (ls <= lt && rt <= rs)
{
add(rot, x, y);
return;
}
int mid = (lt + rt) >> 1;
if (ls <= mid)
insert(rot << 1, lt, mid, ls, rs, x, y);
if (rs > mid)
insert(rot << 1 | 1, mid + 1, rt, ls, rs, x, y);
}
inline void solve(const int rot, const int lt, const int rt)
{
int bck = UFS::top;
for (int i = head[rot]; ~i; i = e[i].next)
UFS::merge(e[i].x, e[i].y);
if (lt == rt)
{
if (arr[lt].type == QUERY)
ans[lt] = (ll)UFS::query(arr[lt].x) * UFS::query(arr[lt].y);
UFS::undo(bck);
return;
}
int mid = (lt + rt) >> 1;
solve(rot << 1, lt, mid);
solve(rot << 1 | 1, mid + 1, rt);
UFS::undo(bck);
}
}
map<pair<int, int>, int> lastins;
int work()
{
read(n), read(q);
UFS::init(n);
Segment_Tree::init();
for (int i = 1; i <= q; i++)
{
char opt;
read(opt), read(arr[i].x), read(arr[i].y);
if (arr[i].x > arr[i].y)
swap(arr[i].x, arr[i].y);
arr[i].type = (opt == 'Q' ? QUERY : ADD);
pair<int, int> p = make_pair(arr[i].x, arr[i].y);
if (arr[i].type == ADD)
lastins[p] = i;
else
{
Segment_Tree::insert(1, 1, q, lastins[p], i - 1, p.first, p.second);
lastins[p] = i + 1;
}
}
for (map<pair<int, int>, int>::iterator it = lastins.begin(); it != lastins.end(); it++)
if (it->second <= q)
Segment_Tree::insert(1, 1, q, it->second, q, it->first.first, it->first.second);
Segment_Tree::solve(1, 1, q);
for (int i = 1; i <= q; i++)
if (arr[i].type == QUERY)
write(ans[i]), putchar('\n');
return (0^_^0);
}
}
int main()
{
return zyt::work();
}
