【bzoj3295】动态逆序对 CDQ分治

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AC通道：http://www.lydsy.com/JudgeOnline/problem.php?id=3295

【题解】

这题很神。

我用的是popoqqq大爷的做法。

具体见http://blog.csdn.net/popoqqq/article/details/38761287

感觉这种做法似乎应该称为整体二分？

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<ctime>#include<algorithm>using namespace std;typedef long long ll;#define FILE "read"#define MAXN 100010#define up(i,j,n) for(int i=j;i<=n;++i)#define dn(i,j,n) for(int i=j;i>=n;--i)#define cmax(a,b) a=max(a,b)#define cmin(a,b) a=min(a,b)struct node{int x,y,pos;bool operator < (const node &b) const {return y<b.y;}}q[MAXN],stack[MAXN];int n,m,tot,a[MAXN],b[MAXN],cnt[MAXN],tr[MAXN],f[MAXN],tim[MAXN];ll ans;namespace INIT{char buf[1<<15],*fs,*ft;inline char getc(){return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++;}inline int read(){int x=0,f=1;  char ch=getchar();while(!isdigit(ch))  {if(ch=='-')  f=-1;  ch=getchar();}while(isdigit(ch))  {x=x*10+ch-'0';  ch=getchar();}return x*f;}}using namespace INIT;bool cmp(node a,node b) {return a.pos<b.pos;}void updata(int x,int f) {for(;x&&x<=n;x+=(x&-x)*f){if(tim[x]!=tot)tr[x]=0,tim[x]=tot;tr[x]++;}}int get(int x,int f){int sum(0);for(;x&&x<=n;x+=(x&-x)*f)if(tim[x]==tot)sum+=tr[x];return sum;}void CDQ(int l,int r){if(l>=r)  return;int mid=(l+r)>>1,ta(l),tb(mid+1),j(l);up(i,l,r) stack[(q[i].pos<=mid?ta:tb)++]=q[i];up(i,l,r) q[i]=stack[i];CDQ(l,mid);  tot++;up(i,mid+1,r){for(;j<=mid&&q[j].y<q[i].y;++j) updata(q[j].x,-1);f[q[i].pos]+=get(q[i].x,1);}tot++; j=mid;dn(i,r,mid+1){for(;j>=l&&q[j].y>q[i].y;--j) updata(q[j].x,1);f[q[i].pos]+=get(q[i].x,-1);}CDQ(mid+1,r);ta=l;tb=mid+1;      up(i,l,r) if((q[ta]<q[tb]||tb>r)&&ta<=mid) stack[i]=q[ta++];  else stack[i]=q[tb++];  up(i,l,r)  q[i]=stack[i];}int main(){freopen(FILE".in","r",stdin);freopen(FILE".out","w",stdout);n=read();  m=read();up(i,1,n)  a[i]=read(),b[a[i]]=i;up(i,1,n){cnt[i]=get(a[i],1);updata(a[i],-1);ans+=cnt[i];}tot++;dn(i,n,1){cnt[i]+=get(a[i],-1);updata(a[i],1);}up(i,1,m)q[i].x=read(),q[i].y=b[q[i].x],q[i].pos=i;sort(q+1,q+m+1);  CDQ(1,m);sort(q+1,q+m+1,cmp);up(i,1,m){printf("%lld\n",ans);ans-=cnt[q[i].y];ans+=f[i];}return 0;}

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