Luogu P3939 数颜色

题目传送门

数据结构学傻了……

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对每个颜色开一个vector，记录该颜色出现过的位置
查询操作直接在vector里二分查找\(l,r\)，一减就可以
修改操作直接二分后swap就可以了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define LL long long
using namespace std;
LL read() {
	LL k = 0, f = 1; char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9')
		k = k * 10 + c - 48, c = getchar();
	return k * f;
}
vector <int> col[300010];
int a[300010];
int main() {
	int n = read(), q = read();
	for(int i = 1; i <= n; ++i)
		a[i] = read(), col[a[i]].push_back(i);
	while(q--) {
		int opt = read();
		if(opt == 1) {
			int l = read(), r = read(), c = read();
			int x = lower_bound(col[c].begin(), col[c].end(), l) - col[c].begin() - 1;
			int y = upper_bound(col[c].begin(), col[c].end(), r) - col[c].begin() - 1;
			if(y-x < 0) printf("0\n");
			else printf("%d\n", y-x);
		}
		else {
			int pos = read();
			if(a[pos] == a[pos+1]) continue;
			int x = lower_bound(col[a[pos]].begin(), col[a[pos]].end(), pos) - col[a[pos]].begin();
			int y = lower_bound(col[a[pos+1]].begin(), col[a[pos+1]].end(), pos+1) - col[a[pos+1]].begin();
			swap(col[a[pos]][x], col[a[pos+1]][y]); swap(a[pos], a[pos+1]);
		}
	}
	return 0;
}