BZOJ 3236: [Ahoi2013]作业(莫队+树状数组)

传送门

解题思路

　　莫队+树状数组。把求\([a,b]\)搞成前缀和形式，剩下的比较裸吧，用\(cnt\)记一下数字出现次数。时间复杂度\(O(msqrt(n)log(n)\)，莫名其妙过了。

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>

using namespace std;
const int MAXM = 1000005;
const int MAXN = 100005;

inline int rd(){
	int x=0,f=1;char ch=getchar();
	while(!isdigit(ch)) f=ch=='-'?0:1,ch=getchar();
	while(isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	return f?x:-x;
}

int n,m,ans[MAXM][2],a[MAXN],f[MAXM*3],g[MAXM*3],siz,cnt[MAXM*3];
int cpy[MAXM*3],num;

struct Data{
	int l,r,a,b,id;
	friend bool operator<(Data A,Data B){
		if(A.l/siz!=B.l/siz) return A.l<B.l;
		if((A.l/siz)&1) return A.r>B.r;
		return A.r<B.r;
	}
}data[MAXM];

inline void add1(int x,int k){
	for(;x<=n;x+=x&-x) f[x]+=k;
}
inline void add2(int x,int k){
	for(;x<=n;x+=x&-x) g[x]+=k;
}
inline int query1(int x){
	if(x<1) return 0;int ret=0;
	for(;x;x-=x&-x) ret+=f[x];
	return ret;
}
inline int query2(int x){
	if(x<1) return 0;int ret=0;
	for(;x;x-=x&-x) ret+=g[x];
	return ret;
}

int main(){
	n=rd(),m=rd();siz=sqrt(n)+1;
	for(int i=1;i<=n;i++) a[i]=rd(),cpy[++num]=a[i];
	for(int i=1;i<=m;i++){
		data[i].l=rd(),data[i].r=rd(),data[i].a=rd();
		data[i].b=rd(),data[i].id=i;cpy[++num]=data[i].a;
		cpy[++num]=data[i].b;
	}
	sort(cpy+1,cpy+1+num);int u=unique(cpy+1,cpy+1+num)-cpy-1;
	for(int i=1;i<=n;i++) a[i]=lower_bound(cpy+1,cpy+1+u,a[i])-cpy;
	for(int i=1;i<=m;i++){
		data[i].a=lower_bound(cpy+1,cpy+1+u,data[i].a)-cpy;
		data[i].b=lower_bound(cpy+1,cpy+1+u,data[i].b)-cpy;
	}
	sort(data+1,data+1+m);
	int L=1,R=0,l,r;
	for(int i=1;i<=m;i++){
		l=data[i].l,r=data[i].r;
		while(R>r) {add1(a[R],-1);if(cnt[a[R]]==1) add2(a[R],-1);cnt[a[R]]--;R--;}
		while(R<r) {R++;add1(a[R],1);if(!cnt[a[R]]) add2(a[R],1);cnt[a[R]]++;}
		while(L>l) {L--;add1(a[L],1);if(!cnt[a[L]]) add2(a[L],1);cnt[a[L]]++;}
		while(L<l) {add1(a[L],-1);if(cnt[a[L]]==1) add2(a[L],-1);cnt[a[L]]--;L++;}
		ans[data[i].id][0]=query1(data[i].b)-query1(data[i].a-1);
		ans[data[i].id][1]=query2(data[i].b)-query2(data[i].a-1);
	}
	for(int i=1;i<=m;i++)
		printf("%d %d\n",ans[i][0],ans[i][1]);
	return 0;
}