HDU 1394(逆序数)

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 14950    Accepted Submission(s): 9138

Problem 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

 

//逆序数的应用

//树状数组#include <iostream>#include <stdio.h>#include <string.h>#include <cmath>#include <algorithm>#include <map>#include <queue>using namespace std;const int maxn=5000+10;const int inf=(1<<30);int n,c[maxn];int arr[maxn];int lowbit(int x){    return x&(-x);}void add(int i,int val){    while(i<=n)    {        c[i]+=val;        i+=lowbit(i);    }}int sum(int i){    int sum=0;    while(i)    {        sum+=c[i];        i-=lowbit(i);    }    return sum;}int main(){    while(~scanf("%d",&n))    {        int ans=0;        memset(c,0,sizeof(c));        for(int i=1;i<=n;i++)        {            scanf("%d",arr+i);            add(arr[i]+1,1);            ans+=(i-sum(arr[i]+1));        }        int sum=ans;        for(int i=1;i<=n;i++)        {            sum-=arr[i];            sum+=n-arr[i]-1;  // res=t-sum(arr[i])+n-(arr[i]+1); 减去小于它的个数加上大于它的个数            ans=min(ans,sum);        }        printf("%d\n",ans);    }    return 0;}//线段树#include <iostream>#include <stdio.h>#include <string.h>#include <cmath>#include <algorithm>#include <map>#include <queue>using namespace std;const int maxn=5000+10;const int inf=(1<<30);int num;int arr[maxn];struct Node{    int r,l;    int num;    int mid()    {        return (r+l)/2;    }}tree[maxn];void Buildtree(int rt,int l,int r){    tree[rt].l=l;    tree[rt].r=r;    tree[rt].num=0;    if(l!=r)    {        Buildtree(2*rt,l,(l+r)/2);        Buildtree(2*rt+1,(l+r)/2+1,r);    }}void Update(int rt,int num,int val){    if(tree[rt].l==tree[rt].r&&tree[rt].r==num)    {        tree[rt].num=val;        return;    }    if(num<=tree[rt].mid())        Update(2*rt,num,val);    else        Update(2*rt+1,num,val);    tree[rt].num=tree[rt*2].num+tree[rt*2+1].num;}void Query(int rt,int l,int r){    if(tree[rt].l==l&&tree[rt].r==r)    {        num+=tree[rt].num;        return;    }    if(r<=tree[rt].mid())        Query(2*rt,l,r);    else if(l>tree[rt].mid())        Query(2*rt+1,l,r);    else    {        Query(2*rt,l,tree[rt].mid());        Query(2*rt+1,tree[rt].mid()+1,r);    }}int main(){    int n;    while(~scanf("%d",&n))    {        Buildtree(1,0,n-1);        int sum=0;        for(int i=1;i<=n;i++)        {            num=0;            scanf("%d",arr+i);            Query(1,arr[i],n-1);            sum+=num;            Update(1,arr[i],1);        }        int res=0;        int ans=sum;        for(int i=1;i<=n;i++)        {            sum-=arr[i];            sum+=n-arr[i]-1;            ans=min(ans,sum);        }        printf("%d\n",ans);    }    return 0;}