acwing 2154. 梦幻布丁

给一个区间，有(n)数，每个数有一个颜色。
有(2)种操作

• 把所有的(x)都修改为(y)
• 求有几个颜色段

可以知道计算出最开始有几个颜色段后，对于某个数字的修改，主需要查看这个数字的左边和右边值是否是(y)即可。
那么，进行启发式合并，把一个set的值合并到另一个set里面

#include <bits/stdc++.h>
#define ll long long
#define ld long double
#define CASE int Kase = 0; cin >> Kase; for(int kase = 1; kase <= Kase; kase++)
using namespace std;
template<typename T = long long> inline T read() {
    T s = 0, f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while(isdigit(ch)) {s = (s << 3) + (s << 1) + ch - 48; ch = getchar();} 
    return s * f;
}
#ifdef ONLINE_JUDGE
#define qaq(...) ;
#define qwq(c) ;
#else
#define qwq(a, b) for_each(a, b, [=](int x){cerr << x << " ";}), cerr << std::endl
template <typename... T> void qaq(const T &...args) {
    auto &os = std::cerr;
    (void)(int[]){(os << args << " ", 0)...};
    os << std::endl;
}
#endif
const int N = 1e5 + 5, M = 1e6 + 5, MOD = 1e9 + 7, CM = 998244353, INF = 0x3f3f3f3f; const ll linf = 0x7f7f7f7f7f7f7f7f;
set<int> s[M];
int fa[M], a[N], ans = 0;
void merge(int x, int y) {
	for(set<int>::iterator it = s[x].begin(); it != s[x].end(); it++) {
		ans -= (a[*it - 1] == y) + (a[*it + 1] == y);
		s[y].insert(*it);	
	}
	for(set<int>::iterator it = s[x].begin(); it != s[x].end(); it++) {
		a[*it] = y;
	}
	s[x].clear();
}
void solve(int kase){
	int n = read(), m = read();
	for(int i = 1; i <= n; i++) {
		a[i] = read(); 
		s[a[i]].insert(i);
		fa[a[i]] = a[i]; ans += a[i] != a[i - 1];
	}
	for(int i = 1; i <= m; i++) {
		int op = read();
		if(op == 1) {
			int u = read(), v = read();
			if(u == v) continue;
			if(s[fa[u]].size() > s[fa[v]].size()) swap(fa[v], fa[u]);
			merge(fa[u], fa[v]);
		}else printf("%d
", ans);
	}
}
const bool ISFILE = 0, DUO = 0;
int main(){
	srand(time(NULL));
    clock_t start, finish; double totaltime; start = clock();
    if(ISFILE) freopen("/Users/i/Desktop/practice/in.txt", "r", stdin);
    if(DUO) {CASE solve(kase);} else solve(1);
    finish = clock(); 
    qaq("
Time:", (double)(finish - start) / CLOCKS_PER_SEC * 1000, "ms
");
    return 0;
}