離線。將大小離散,然後用莫隊更新樹狀數組和答案就可以了。
代碼:
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 using namespace std; 7 #define N 50010 8 #define lowbit(x) x&-x 9 struct Node{ 10 int l,r,f; 11 }a[N]; 12 struct Ls{ 13 int w,f; 14 }l[N]; 15 int Ans[N],s,i,j,k,n,m,b[N],c[N],A[N],q,L,R,Res,Sum; 16 inline bool Cmp1(Ls a,Ls b){return a.w<b.w;} 17 inline bool Cmp2(Node x,Node y){ 18 if(b[x.l]==b[y.l])return x.r<y.r; 19 return b[x.l]<b[y.l]; 20 } 21 inline int Query(int x){ 22 int Ans=0; 23 for(;x;x-=lowbit(x))Ans+=c[x]; 24 return Ans; 25 } 26 inline void Update(int x,int y){for(;x<=m;x+=lowbit(x))c[x]+=y;} 27 int main(){ 28 scanf("%d",&n); 29 for(i=1;i<=n;i++)scanf("%d",&l[i].w),l[i].f=i; 30 sort(l+1,l+n+1,Cmp1); 31 A[l[1].f]=m=1; 32 for(i=2;i<=n;i++) 33 if(l[i].w==l[i-1].w)A[l[i].f]=m;else A[l[i].f]=++m; 34 s=sqrt((double)n); 35 for(i=1;i<=n;i++)b[i]=(i-1)/s+1; 36 scanf("%d",&q); 37 for(i=1;i<=q;i++)scanf("%d%d",&a[i].l,&a[i].r),a[i].f=i; 38 sort(a+1,a+q+1,Cmp2); 39 for(i=L=1;i<=q;i++){ 40 while(a[i].l<L)L--,Res+=Query(A[L]-1),Update(A[L],1); 41 while(a[i].r>R)R++,Update(A[R],1),Res+=R-L+1-Query(A[R]); 42 while(a[i].l>L)Res-=Query(A[L]-1),Update(A[L],-1),L++; 43 while(a[i].r<R)Update(A[R],-1),Res-=R-L-Query(A[R]),R--; 44 Ans[a[i].f]=Res; 45 } 46 for(i=1;i<=q;i++)printf("%d\n",Ans[i]); 47 return 0; 48 }bzoj3289