CF848C Goodbye Souvenir

题目描述

Solution

考虑拆贡献，把最后一次的下标减去第一次的下标的和拆成每一个点与和它数字相同的上一个点的差的和，也就是$\sum i-pre[i]$。

这样转化之后，每一次询问一个区间$[l,r]$，相当于找到所有满足条件的$i$，使得$l\le i\le r$，且$pre[i]\ge l$，即$Ans={\sum }_{l\le i\le r,l\le pre[i]}i-pre[i]$，这样单个询问就是一个简单的二维数点问题了（事实上只需要满足$i\le r,l\le pre[i]$即可，因为$pre[i]$始终小于$i$）。

再思考如何修改，直接$CDQ$分治即可，相当于加上一个时间轴，类似一个三维数点。
时间复杂度$O(nl{g}^{2}n)$。

Code

代码不知道对不对，因为$luogu$交完过了$7$个点之后$UKE$了？！？

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
ll Ans[MAXN];
set<int> S[MAXN];
int pre[MAXN],nxt[MAXN],whp[MAXN],whn[MAXN],a[MAXN],n,m,num=0,Num=0;
struct Qnode{ int id,opt,x,y,c; } Q[MAXN];
int compare(Qnode x,Qnode y)
{if (x.opt!=y.opt) return x.opt<y.opt;return x.opt?(x.y<y.y):(x.x<y.x);
}
int compareid(Qnode x,Qnode y) { return x.id<y.id; }
void Init()
{n=read(),m=read();for (int i=1;i<=n;i++) a[i]=read();for (int i=1;i<=n;i++) whp[i]=0,whn[i]=n+1;for (int i=1;i<=n;i++) pre[i]=whp[a[i]],whp[a[i]]=i;for (int i=n;i>=1;i--) nxt[i]=whn[a[i]],whn[a[i]]=i;for (int i=1;i<=n;i++) Q[++num]=(Qnode){num,0,i,pre[i],i-pre[i]},S[a[i]].insert(i);while (m--){int opt=read(),y=read(),z=read();if (opt==1) {if (pre[y]!=0)   nxt[pre[y]]=nxt[y],Q[++num]=(Qnode){num,0,y,pre[y],pre[y]-y};if (nxt[y]!=n+1) pre[nxt[y]]=pre[y],Q[++num]=(Qnode){num,0,nxt[y],y,y-nxt[y]};if (pre[y]!=0&&nxt[y]!=n+1) Q[++num]=(Qnode){num,0,nxt[y],pre[y],nxt[y]-pre[y]};S[a[y]].erase(y),a[y]=z;set<int>::iterator it=S[z].lower_bound(y);nxt[y]=(it!=S[z].end())?(*it):n+1;pre[y]=(it!=S[z].begin())?(*(--it)):0;if (pre[y]!=0)   nxt[pre[y]]=y,Q[++num]=(Qnode){num,0,y,pre[y],y-pre[y]};if (nxt[y]!=n+1) pre[nxt[y]]=y,Q[++num]=(Qnode){num,0,nxt[y],y,nxt[y]-y};if (pre[y]!=0&&nxt[y]!=n+1) Q[++num]=(Qnode){num,0,nxt[y],pre[y],pre[y]-nxt[y]};		S[z].insert(y);}else Q[++num]=(Qnode){num,1,y,z,++Num};}
//	for (int i=1;i<=num;i++) cout<<Q[i].id<<" "<<Q[i].opt<<" "<<Q[i].x<<" "<<Q[i].y<<" "<<Q[i].c<<endl;
}
struct Binary_Index_Tree
{ll s[MAXN<<1];int lowbit(int x) { return x&(-x); }void change(int x,int y){ if (!x) return; for (;x<=n;x+=lowbit(x)) s[x]+=y; }ll query(int x) { ll ans=0; for (;x;x-=lowbit(x)) ans+=s[x]; return ans; }
} bit;
void Solve(int l,int r)
{if (l==r) return;int mid=(l+r)>>1;Solve(l,mid),Solve(mid+1,r);sort(Q+l,Q+mid+1,compare);sort(Q+mid+1,Q+r+1,compare);int L=l;for (int i=mid+1;i<=r;i++)if (Q[i].opt){while (L<=mid&&(Q[L].opt||Q[L].x<=Q[i].y)) { if (!Q[L].opt) bit.change(Q[L].y,Q[L].c); L++; }Ans[Q[i].c]+=bit.query(n)-bit.query(Q[i].x-1);}for (int i=l;i<L;i++) if (!Q[i].opt) bit.change(Q[i].y,-Q[i].c);
}
int main()
{Init();Solve(1,num);for (int i=1;i<=Num;i++) printf("%lld\n",Ans[i]);return 0;
}