hdu 1394 线段树

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16046    Accepted Submission(s): 9763

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

/*hdu 1394 求a[i]前面小于a[i]的数的个数给你一个序列，求每个a[i]后面小于a[i]的个数，然后你可以把第一个数放到最后，这样的话sum变化：sum = sum+(n-1-a[i])-a[i];hhh-2016-02-27 15:18:09*/#include <iostream>#include <cstdio>#include <cstring>#include <ctime>#include <algorithm>#include <cmath>#include <queue>#include <map>#include <vector>typedef long long ll;using namespace std;const int maxn = 200000+5;int a[maxn];struct node{    int l,r;    int num;} tree[maxn<<2];void push_up(int r){    int lson = r<<1,rson = (r<<1)|1;    tree[r].num = (tree[lson].num+tree[rson].num);}void build(int i,int l,int r){    tree[i].l = l,tree[i].r = r;    tree[i].num  = 0;    if(l == r)    {        return ;    }    int mid = (l+r)>>1;    build(i<<1,l,mid);    build(i<<1|1,mid+1,r);    push_up(i);}void push_down(int r){}void Insert(int i,int k){    if(tree[i].l == k && tree[i].r == k)    {        tree[i].num++;        return ;    }    push_down(i);    int mid = (tree[i].l + tree[i].r) >>1;    if(k <= mid) Insert(i<<1,k);    if(k > mid) Insert(i<<1|1,k);    push_up(i);}int query(int i,int l,int r){    if(tree[i].l >= l && tree[i].r <= r)    {        return tree[i].num;    }    push_down(i);    int mid = (tree[i].l+tree[i].r)>>1;    int ans = 0;    if(l <= mid) ans+=(query(i<<1,l,r));    if(r > mid) ans+=(query(i<<1|1,l,r));    return ans ;}int main(){    int T,n,m,cas = 1;    while(scanf("%d",&n)!=EOF)    {        build(1,0,n-1);        int sum = 0;        for(int i =1; i <= n; i++)        {            scanf("%d",&a[i]);            Insert(1,a[i]);            int t;            if(a[i] - 1 < 0)                t = 0;            else               t = query(1,0,a[i]-1);            sum += (a[i] - t);        }        int ans = sum;        for(int i = 1;i <= n;i++)        {            sum = sum+(n-1-a[i])-a[i];            ans = min(ans,sum);        }        printf("%d\n",ans);    }    return 0;}