[BZOJ]4991: [Usaco2017 Feb]Why Did the Cow Cross the Road III CDQ分治+树状数组

Description

Farmer John is continuing to ponder the issue of cows crossing the road through his farm, introduced in the preceding two problems. He realizes now that the threshold for friendliness is a bit more subtle than he previously considered – breeds aa and bb are now friendly if |a-b|≤K, and unfriendly otherwise.Given the orderings of fields on either side of the road through FJ’s farm, please count t
he number of unfriendly crossing pairs of breeds, where a crossing pair of breeds is defined as in the preceding problems.

题解：

首先题目大意：两列n的排列，相同的数连边，假如一对数的差>k且有交叉，ans++，求ans。对于每个数有三个属性，一个是在第一个序列中的位置，一个是在第二个序列中的位置，一个是它的值，就是个三维偏序问题，简单的CDQ分治，树套树空间过不去。

代码：

#include<bits/stdc++.h>using namespace std;#define LL long long#define pa pair<int,int>const int Maxn=100010;const int inf=2147483647;int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+(ch^48),ch=getchar();    return x*f;}int n,k,a[Maxn],b[Maxn],posa[Maxn],posb[Maxn];struct Opt{int x,y,z;}q[Maxn],temp[Maxn];bool cmp(Opt a,Opt b){return a.x>b.x;}int s[Maxn];void add(int x,int y){for(;x<=n;x+=(x&-x))s[x]+=y;}int getsum(int x){int re=0;x=min(x,n);for(;x>0;x-=(x&-x))re+=s[x];return re;}LL ans=0;void solve(int l,int r){    if(l==r)return;    int mid=l+r>>1;    solve(l,mid);solve(mid+1,r);    int i=l,j=mid+1,len=0;    while(i<=mid&&j<=r)    {        if(q[i].y<q[j].y)        {            add(q[i].z,1);            temp[++len]=q[i++];        }        else        {            ans+=(LL)(getsum(q[j].z-k-1))+(LL)(getsum(n)-getsum(q[j].z+k));            temp[++len]=q[j++];        }    }    int t=i;    while(i<=mid)temp[++len]=q[i++];    while(j<=r)ans+=(LL)(getsum(q[j].z-k-1))+(LL)(getsum(n)-getsum(q[j].z+k)),temp[++len]=q[j++];    for(int p=l;p<t;p++)add(q[p].z,-1);    for(int p=1;p<=len;p++)q[p+l-1]=temp[p];}int main(){    n=read(),k=read();    for(int i=1;i<=n;i++)a[i]=read(),posa[a[i]]=i;    for(int i=1;i<=n;i++)b[i]=read(),posb[b[i]]=i;    for(int i=1;i<=n;i++)q[i].x=posa[i],q[i].y=posb[i],q[i].z=i;    sort(q+1,q+1+n,cmp);solve(1,n);    printf("%lld",ans);}