题目：
分析：
（这题码了一下午，码了近250行，但是意外跑的比本校各位神仙稍快，特写博客纪念）
首先能看出一个显然的结论：颜色段数只会变少不会变多。
我们考虑用并查集维护区间，对于每个区间维护它的起点和终点。建(n)棵平衡树，第(i)棵存颜色为(i)的区间。把(x)变成(y)时进行启发式合并，同时对于(x)上的每个结点([a,b])，在(y)中找(a-1)和(b+1)所在区间。如果存在则合并，并答案减(1)；若不存在则向(y)中插入新结点。时间复杂度(O(mlog ^2n))
代码：
代码不难写，就是长……
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;
namespace zyt
{
template<typename T>
inline void read(T &x)
{
char c;
bool f = false;
x = 0;
do
c = getchar();
while (c != '-' && !isdigit(c));
if (c == '-')
f = true, c = getchar();
do
x = x * 10 + c - '0', c = getchar();
while (isdigit(c));
if (f)
x = -x;
}
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);
}
const int N = 1e6 + 10;
int n, m, p[N], st[N], ed[N];
int f(const int x)
{
return x == p[x] ? x : p[x] = f(p[x]);
}
int ans = 0;
class Splay
{
private:
struct node
{
int val, size;
node *fa, *s[2];
node(const int _val, node *_fa)
: val(_val), fa(_fa)
{
size = 1;
s[0] = s[1] = NULL;
}
}*head;
bool dir(const node *rot)
{
return rot == rot->fa->s[1];
}
void update(node *rot)
{
rot->size = 1;
if (rot->s[0])
rot->size += rot->s[0]->size;
if (rot->s[1])
rot->size += rot->s[1]->size;
}
void rotate(node *rot)
{
node *f = rot->fa, *ff = f->fa;
bool d = dir(rot);
rot->fa = ff;
if (ff)
ff->s[dir(f)] = rot;
else
head = rot;
f->s[d] = rot->s[!d];
if (rot->s[!d])
rot->s[!d]->fa = f;
rot->s[!d] = f;
f->fa = rot;
update(f);
}
void splay(node *rot, const node *goal = NULL)
{
while (rot && rot->fa && rot->fa != goal)
{
node *f = rot->fa, *ff = f->fa;
if (ff == goal)
rotate(rot);
else if (dir(rot) ^ dir(f))
rotate(rot), rotate(rot);
else
rotate(f), rotate(rot);
}
update(rot);
}
void del(node *rot, Splay &s)
{
if (!rot)
return;
int x = f(rot->val);
bool flag = false;
if (st[x] > 1 && s.find(f(st[x] - 1)))
{
p[x] = f(st[x] - 1);
ed[f(st[x] - 1)] = ed[x];
--ans, flag = true;
}
x = f(x);
if (ed[x] < n && s.find(f(ed[x] + 1)))
{
p[x] = f(ed[x] + 1);
st[f(ed[x] + 1)] = st[x];
--ans, flag = true;
}
if (!flag)
s.insert(rot->val);
if (rot->s[0])
del(rot->s[0], s);
if (rot->s[1])
del(rot->s[1], s);
delete rot;
}
node *find(const int val)
{
node *rot = head;
while (1)
{
if (!rot || rot->val == val)
return rot;
if (val < rot->val)
rot = rot->s[0];
else
rot = rot->s[1];
}
}
public:
void insert(const int val)
{
if (!head)
{
head = new node(val, NULL);
return;
}
node *rot = head;
while (1)
{
if (val < rot->val)
{
if (rot->s[0])
rot = rot->s[0];
else
{
rot->s[0] = new node(val, rot);
splay(rot->s[0]);
break;
}
}
else
{
if (rot->s[1])
rot = rot->s[1];
else
{
rot->s[1] = new node(val, rot);
splay(rot->s[1]);
break;
}
}
}
}
size_t size()
{
if (head)
return head->size;
else
return 0;
}
friend void merge(Splay &a, Splay &b);
}tree[N];
inline void merge(Splay &a, Splay &b)
{
if (a.size() < b.size())
swap(a, b);
b.del(b.head, a);
b.head = NULL;
}
int arr[N], tmp[N];
int work()
{
read(n), read(m);
int last = 1;
for (int i = 1; i <= n; i++)
read(arr[i]);
p[1] = st[1] = ed[1] = 1;
for (int i = 2; i <= n; i++)
{
if (arr[i] != arr[i - 1])
{
tree[arr[i - 1]].insert(last);
st[i] = last = i, ++ans;
}
p[i] = last;
ed[last] = i;
}
tree[arr[n]].insert(last), ++ans;
while (m--)
{
int opt;
read(opt);
if (opt == 1)
{
int x, y;
read(x), read(y);
if (x != y)
merge(tree[y], tree[x]);
}
else
write(ans), putchar('
');
}
return 0;
}
}
int main()
{
return zyt::work();
}