[Luogu 1908] Reverse Order, luogu1908 Reverse Order

[Luogu 1908] Reverse Order, luogu1908 Reverse Order
Description

TOM and JERRY, the cat and CAT, have been competing again recently, but after all they are adults. They no longer like to play the game that you catch up with. Now they like to play statistics. Recently, TOM found something that humans call "backward-order pairs". This is defined as follows: for a given positive integer sequence, reverse Order pairs are the ordered pairs of ai> aj and I <j in the sequence. After knowing this concept, they will first calculate the number of reverse pairs in a given positive integer sequence.

Input/Output Format

Input Format:

The first row, n, indicates the number of n in the sequence.

N number in the second row, indicating the given sequence.

Output Format:

Number of reverse pairs in a given sequence.

Input and Output sample input sample #1:

65 4 2 6 3 1

Output sample #1:

11

Description

For 50% of data, n ≤ 2500

For 100% of data, n ≤ 40000.

Tree array (to be discretization ):

 1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<algorithm> 5 #define maxn 40005 6 using namespace std; 7 int read(){ 8     int x=0,f=1;char ch=getchar(); 9     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}10     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}11     return x*f;12 }13 int n,ans=0,c[maxn],b[maxn];14 int lowbit(int x){return x&(-x);}15 void add(int pos,int x){16     while(pos<=n){17         c[pos]+=x;18         pos+=lowbit(pos);19     }20 }21 int getsum(int pos){22     int ans=0;23     while(pos>0){24         ans+=c[pos];25         pos-=lowbit(pos);26     }27     return ans;28 }29 struct node{30     int w,id;31     bool operator < (const node &a) const{32         return w<a.w;33     }34 }a[maxn];35 int main(){36     n=read();37     for(int i=1;i<=n;i++){38         a[i].w=read();39         a[i].id=i;40     }41     sort(a+1,a+n+1);42     int cnt=0;43     b[a[1].id]=++cnt;44     for(int i=2;i<=n;i++){45         if(a[i-1].w!=a[i].w) b[a[i].id]=++cnt;46         else b[a[i].id]=cnt;47     }48     for(int i=1;i<=n;i++){49         add(b[i],1);50         ans+=(getsum(n)-getsum(b[i]));51     }52     printf("%d",ans);53     return 0;54 }

Merge Sorting:

 1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cstring> 5 using namespace std; 6 int read(){ 7     int x=0,f=1;char ch=getchar(); 8     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 9     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}10     return x*f;11 }12 int n,a[40004],rr[40004],ans=0;13 void merge(int l,int r){14     if(l==r) return ;15     int mid=(l+r)>>1;merge(l,mid);merge(mid+1,r);16     int t1=l,t2=mid+1,cnt=0;17     while(t1<=mid&&t2<=r){18         if(a[t1]<a[t2])19             rr[++cnt]=a[t1],t1++;20         else{21             rr[++cnt]=a[t2],t2++;22             ans+=mid-t1+1;23         }24     }25     while(t1<=mid) rr[++cnt]=a[t1],t1++;26     while(t2<=mid) rr[++cnt]=a[t2],t2++;27     for(int i=1;i<=cnt;i++) a[l+i-1]=rr[i];28 }29 int main(){30     n=read();for(int i=1;i<=n;i++)a[i]=read();31     merge(1,n);32     printf("%d",ans);33     return 0;34 }

Line Segment tree (to be discretization ):

 1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<algorithm> 5 #define maxn 40005 6 #define ll l,mid,rt<<1 7 #define rr mid+1,r,rt<<1|1 8 using namespace std; 9 int read(){10     int x=0,f=1;char ch=getchar();11     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}12     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}13     return x*f;14 }15 int sum[maxn<<2],b[maxn<<2];16 struct node{17     int w,id;18     bool operator < (const node &a) const{19         return w<a.w;20     }21 }a[maxn<<2];22 void build(int l,int r,int rt){23     if(l==r){24         sum[rt]=0;25         return ;26     }27     int mid=(l+r)>>1;28     build(ll);build(rr);29     sum[rt]=sum[rt<<1]+sum[rt<<1|1];30 }31 int query(int ql,int qr,int l,int r,int rt){32     if(ql<=l&&r<=qr)33         return sum[rt];34     int mid=(l+r)>>1;35     int ans=0;36     if(ql<=mid) ans+=query(ql,qr,ll);37     if(mid<qr) ans+=query(ql,qr,rr);38     return ans;39 }40 void update(int pos,int l,int r,int rt){41     if(l==r){42         sum[rt]=1;43         return ;44     }45     int mid=(l+r)>>1;46     if(pos<=mid) update(pos,ll);47     else update(pos,rr);48     sum[rt]=sum[rt<<1]+sum[rt<<1|1];49 }50 int main(){51     int n;52     n=read();53     for(int i=1;i<=n;i++){54         a[i].w=read();55         a[i].id=i;56     }57     sort(a+1,a+n+1);58     int cnt=0;59     b[a[1].id]=++cnt;60     for(int i=2;i<=n;i++){61         if(a[i-1].w!=a[i].w) b[a[i].id]=++cnt;62         else b[a[i].id]=cnt;63     }64     build(1,n,1);65     int ans=0;66     for(int i=1;i<=n;i++){67         ans+=query(b[i],n,1,n,1);68         update(b[i],1,n,1);69     }70     printf("%d",ans);71     return 0;72 }