[HNOI2009]梦幻布丁 BZOJ1483 set

题目描述

N个布丁摆成一行,进行M次操作.每次将某个颜色的布丁全部变成另一种颜色的,然后再询问当前一共有多少段颜色.例如颜色分别为1,2,2,1的四个布丁一共有3段颜色.

输入输出格式

输入格式：

第一行给出N,M表示布丁的个数和好友的操作次数.
第二行N个数A1,A2...An表示第i个布丁的颜色从第三行起有M行,对于每个操作,若第一个数字是1表示要对颜色进行改变，其后的两个整数X,Y表示将所有颜色为X的变为Y，X可能等于Y.
若第一个数字为2表示要进行询问当前有多少段颜色，这时你应该输出一个整数. 0

输出格式：

针对第二类操作即询问，依次输出当前有多少段颜色.

输入输出样例

输入样例#1：
复制

4 3
1 2 2 1
2
1 2 1
2

输出样例#1： 复制

3
1

说明

1<=n,m<=100,000; 0<Ai,x,y<1,000,000

用 set 进行暴力处理；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

ll qpow(ll a, ll b, ll c) {
	ll ans = 1;
	a = a % c;
	while (b) {
		if (b % 2)ans = ans * a%c;
		b /= 2; a = a * a%c;
	}
	return ans;
}

int n, m, ans;
int fath[maxn], v[maxn];
set<int>st[maxn];

void sol(int a, int b) {
	set<int>::iterator it;
	for (it = st[a].begin(); it != st[a].end(); it++) {
		if (v[*it - 1] == b)ans--;
		if (v[*it + 1] == b)ans--;
		st[b].insert(*it);
	}
	for (it = st[a].begin(); it != st[a].end(); it++) {
		v[*it] = b;// 修改为 b
	}
	st[a].clear();
}

int main()
{
	//ios::sync_with_stdio(0);
	rdint(n); rdint(m);
	for (int i = 1; i <= n; i++)rdint(v[i]);
	for (int i = 1; i <= n; i++) {
		fath[v[i]] = v[i];
		if (v[i] != v[i - 1])ans++;
		st[v[i]].insert(i);
	}
	for (int i = 1; i <= m; i++) {
		int op; rdint(op);
		int a, b;
		if (op == 2)cout << ans << endl;
		else {
			rdint(a); rdint(b);
			if (a == b)continue;
			if (st[fath[a]].size() > st[fath[b]].size())swap(fath[a], fath[b]);
			sol(fath[a], fath[b]);
		}
	}
	return 0;
}