【bzoj3295】动态逆序对

 1 #include<bits/stdc++.h>
 2 #define inf 0x7fffffff
 3 #define N 100005
 4 #define M 5000005
 5 using namespace std;
 6 typedef long long ll;
 7 ll ans;
 8 int n,m,sz,a[100],b[100],val[N],pos[N],a1[N],a2[N];
 9 int c[N*10],rt[N],ls[M],rs[M],sumv[M];
10 inline int lowbit(int x){return x&(-x);}
11 inline int ask(int x){
12     int ans=0;
13     for(int i=x;i;i-=lowbit(i))ans+=c[i];
14     return ans;
15 }
16 void change(int &o,int l,int r,int q){
17     if(!o)o=++sz;sumv[o]++;
18     if(l==r)return;
19     int mid=(l+r)>>1;
20     if(q<=mid)change(ls[o],l,mid,q);
21     else change(rs[o],mid+1,r,q);
22 }
23 int querysub(int x,int y,int v){
24     int cnta=0,cntb=0;int ans=0;x--;
25     for(int i=x;i;i-=lowbit(i))a[++cnta]=rt[i];
26     for(int i=y;i;i-=lowbit(i))b[++cntb]=rt[i];
27     int l=1,r=n;
28     while(l!=r){
29         int mid=(l+r)>>1;
30         if(v<=mid){
31             for(int i=1;i<=cnta;i++)ans-=sumv[rs[a[i]]];
32             for(int i=1;i<=cntb;i++)ans+=sumv[rs[b[i]]];
33             for(int i=1;i<=cnta;i++)a[i]=ls[a[i]];
34             for(int i=1;i<=cntb;i++)b[i]=ls[b[i]];
35             r=mid;
36         }
37         else{
38             for(int i=1;i<=cnta;i++)a[i]=rs[a[i]];
39             for(int i=1;i<=cntb;i++)b[i]=rs[b[i]];
40             l=mid+1;
41         }
42     }
43     return ans;
44 }
45 int querypre(int x,int y,int v){
46     int cnta=0,cntb=0,ans=0;x--;
47     for(int i=x;i;i-=lowbit(i))a[++cnta]=rt[i];
48     for(int i=y;i;i-=lowbit(i))b[++cntb]=rt[i];
49     int l=1,r=n;
50     while(l!=r){
51         int mid=(l+r)>>1;
52         if(v>mid){
53             for(int i=1;i<=cnta;i++)ans-=sumv[ls[a[i]]];
54             for(int i=1;i<=cntb;i++)ans+=sumv[ls[b[i]]];
55             for(int i=1;i<=cnta;i++)a[i]=rs[a[i]];
56             for(int i=1;i<=cntb;i++)b[i]=rs[b[i]];
57             l=mid+1;
58         }
59         else{
60             for(int i=1;i<=cnta;i++)a[i]=ls[a[i]];
61             for(int i=1;i<=cntb;i++)b[i]=ls[b[i]];
62             r=mid;
63         }
64     }
65     return ans;
66 }
67 inline int read(){
68     int f=1,x=0;char ch;
69     do{ch=getchar();if(ch==‘-‘)f=-1;}while(ch<‘0‘||ch>‘9‘);
70     do{x=x*10+ch-‘0‘;ch=getchar();}while(ch>=‘0‘&&ch<=‘9‘);
71     return f*x;
72 }
73 int main(){
74     n=read();m=read();
75     for(int i=1;i<=n;i++){
76         val[i]=read();pos[val[i]]=i;
77         a1[i]=ask(n)-ask(val[i]);
78         ans+=a1[i];
79         for(int j=val[i];j<=n;j+=lowbit(j))c[j]++;
80     }
81     memset(c,0,sizeof(c));
82     for(int i=n;i;i--){
83         a2[i]=ask(val[i]-1);
84         for(int j=val[i];j<=n;j+=lowbit(j))c[j]++;
85     }
86     for(int i=1;i<=m;i++){
87         printf("%lld\n",ans);
88         int x=read();x=pos[x];
89         ans-=(a1[x]+a2[x]-querysub(1,x-1,val[x])-querypre(x+1,n,val[x]));
90         for(int j=x;j<=n;j+=lowbit(j))change(rt[j],1,n,val[x]);
91     }
92     return 0;
93 }