【线段树】Minimum Inversion Number（逆序数的求解）

Minimum Inversion Number

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj. 

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following: 

a1, a2, ..., an-1, an (where m = 0 - the initial seqence) 
a2, a3, ..., an, a1 (where m = 1) 
a3, a4, ..., an, a1, a2 (where m = 2) 
... 
an, a1, a2, ..., an-1 (where m = n-1) 

You are asked to write a program to find the minimum inversion number out of the above sequences. 

 

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1. 

 

Output

For each case, output the minimum inversion number on a single line. 

 

Sample Input

 101 3 6 9 0 8 5 7 4 2 

 

Sample Output

 16 

这题用线段树和树状数组都可以做。要注意的有两点，一是数的大小都是0~n-1,所以数状数组和线段树都可以直接把数的大小当成范围，而不是把元素下标当范围。二是每次换序时不用重新算一遍，有规律：ans=ans-a[i]+(n-1-a[i])（即新逆序数的数目=原来逆序数的数目-要移动的数字会减少的逆序数数目+移动后增加的数目），例如把3移动到数组末尾，那么会减少3中逆序数数目（2,1,0），增加（n-1-3）种情况。

树状数组的做法是：每次输入一个数i即将其放入数状数组中，将a[i]和后面相关的加1，然后搜索1~a[i]的总和，这个总和即是比它小的数目，再用i减去总和即为比它大的数目，即逆序数的数目（注意每个输入的数都要加一再放入数状数组）

线段树的做法原理差不多，不多说了。

这题的收获是：1.扩宽了思路，线段树不一定都对应输出数组的下标，可以是数的大小

                             2.了解了逆序数的求解办法，还有一种排序的方法没研究，以后有机会再看吧

两种代码如下：

树状数组：

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>#include<math.h>#include<queue>using namespace std;int n,a[5005],c[5005];int lowbit(int x){    return x&(-x);}void add(int val,int i){    while(i<=n)    {        c[i]+=val;        i+=lowbit(i);    }}int sum(int n){    int ans=0;    while(n)    {        ans+=c[n];        n-=lowbit(n);    }    return ans;}int main(){    int i,j,ans,temp;    while(~scanf("%d",&n))    {        ans=0;        memset(c,0,sizeof(c));        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);            add(1,a[i]+1);            ans+=i-sum(a[i]+1);        }        temp=ans;        for(i=1;i<=n;i++)        {            temp=temp-a[i]+(n-1-a[i]);            if(temp<ans)            ans=temp;        }        cout<<ans<<endl;    }    return 0;}

线段树：

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>#include<math.h>#include<queue>using namespace std;const int n=5005;struct node{    int l,r,num;}t[n<<2];void build(int l,int r,int k){    t[k].l=l;    t[k].r=r;    t[k].num=0;    if(l==r)    return ;    int mid=(l+r)>>1;    build(l,mid,2*k);    build(mid+1,r,2*k+1);}void insert(int num,int k){    if(t[k].l==t[k].r)    {        t[k].num++;     //   cout<<t[k].l<<" "<<t[k].r<<" "<<t[k].num<<endl;        return ;    }    int mid=(t[k].l+t[k].r)>>1;    if(num<=mid)    insert(num,2*k);    else    insert(num,2*k+1);    t[k].num=t[2*k].num+t[2*k+1].num;}int search(int k,int l,int r){   // cout<<"f:"<<k<<" "<<t[k].l<<" "<<t[k].r<<endl;;    if(t[k].l==l&&t[k].r==r)    return t[k].num;    int mid=(t[k].l+t[k].r)>>1;    int ans=0;    if(l>=mid+1)    ans=search(2*k+1,l,r);    else if(r<=mid)    ans=search(2*k,l,r);    else    {        ans+=search(2*k,l,mid);        ans+=search(2*k+1,mid+1,r);    }  //  cout<<"f:"<<t[k].l<<" "<<t[k].r<<" "<<t[k].num<<endl;    return ans;}int main(){    int i,j,ans,temp,N,a[n];    while(~scanf("%d",&N))    {        build(0,N-1,1);        ans=0;        for(i=1;i<=N;i++)        {            scanf("%d",&a[i]);            insert(a[i],1);            ans+=i-search(1,0,a[i]);         //   cout<<"0~"<<a[i]<<" "<<search(1,0,a[i])<<endl;        //    cout<<ans<<endl;        }        temp=ans;      //  cout<<ans<<endl;        for(i=1;i<=N;i++)        {            temp=temp-a[i]+(N-a[i]-1);         //   cout<<temp<<endl;            if(temp<ans)            ans=temp;        }        cout<<ans<<endl;    }    return 0;}