【题解】P3939 数颜色

不要数据结构和模板学傻了...

考虑到兔子们交换都是相邻的，说明任何一次交换只会引起\(O(1)\)的变化。

我们开很多\(vector\)存没种兔子的下标就好了。到时候二分查找查询。

复杂度\(O(nlogn)\)

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<bitset>
#include<vector>
#include<map>
#include<ctime>
#include<cstdlib>
#include<set>
#include<bitset>
#include<stack>
#include<list>
#include<cmath>
using namespace std;
#define RP(t,a,b) for(register int (t)=(a),edd_=(b);t<=edd_;++t)
#define DRP(t,a,b) for(register int (t)=(a),edd_=(b);t>=edd_;--t)
#define ERP(t,a) for(int t=head[a];t;t=e[t].nx)
#define Max(a,b) ((a)<(b)?(b):(a))
#define Min(a,b) ((a)<(b)?(a):(b))
#define TMP template<class ccf>
typedef long long ll;
TMP inline ccf qr(ccf k){char c=getchar();ccf x=0;int q=1;while(c<48||c>57)q=c==45?-1:q,c=getchar();while(c>=48&&c<=57)x=x*10+c-48,c=getchar();if(q==-1)x=-x;return x;
}
const int maxn=3e5+15;
int data[maxn];
vector < int > col[maxn];
int n,m;
inline int div(vector < int >& p,int lb,int rb,int k){// to lbint mid;int ret=0;do{mid=(lb+rb)>>1;if(p[mid]>=k)rb=mid-1,ret=mid;elselb=mid+1;}while(lb<=rb);return ret;
}inline int div2(vector < int > &p,int lb,int rb,int k){int ret=-1;int mid;do{mid=(lb+rb)>>1;if(p[mid]<=k)lb=mid+1,ret=mid;elserb=mid-1;}while(lb<=rb);return ret;
}inline void upd(int c,int pos,int to){int ret=div(col[c],0,col[c].size()-1,pos);if(ret<0)return void(cout<<"err!\n");col[c][ret]=to;
}inline int cnt(int c,int lb,int rb){int ret=0;int siz=col[c].size()-1;if(siz<0)return 0;if(rb<col[c][0])return 0;if(lb>col[c][siz])return 0;ret+=div2(col[c],0,siz,rb);ret-=div(col[c],0,siz,lb);return ret+1;
}int t1,t2,t3,t4;
int main(){
#ifndef ONLINE_JUDGEfreopen("in.in","r",stdin);freopen("out.out","w",stdout);
#endifn=qr(1);m=qr(1);RP(t,1,n){data[t]=qr(1);col[data[t]].push_back(t);}RP(t,1,m){t1=qr(1);if(t1==1){t2=qr(1);t3=qr(1);t4=qr(1);cout<<cnt(t4,t2,t3)<<endl;}else{t2=qr(1);if(data[t2]==data[t2+1])continue;else{upd(data[t2],t2,t2+1);upd(data[t2+1],t2+1,t2);swap(data[t2],data[t2+1]);}}}return 0;
}