hdu 1394 Minimum Inversion Number

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
10
1 3 6 9 0 8 5 7 4 2
Sample Output

16

题意：给一个n个为0~n-1的数，求n个数列 的逆序对数  输出最小的值。

线段树：

刚刚学习线段树，看了大神的提示才懂得，确实有规律的：sum[m=i]  = sum[m=i-1]+(n-2*a[i]-1) ：

 用C[n]的数组和线段树来求前面大于a[i]的数。

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define siz 5005
using namespace std;
int n,C[siz],tree[siz*4],a[siz];
void build(int node,int s,int e)
{
    if(s==e)
        tree[node]=C[s];
    else{
        int mid=(s+e)/2;
        build(node*2,s,mid);
        build(node*2+1,mid+1,e);
        tree[node]=tree[node*2]+tree[node*2+1];
    }
}
void updataone(int node, int s,int e,int v)
{
    if(s==e){
        tree[node]=C[v];
    }else{
        int mid=(s+e)/2;
        if(v<=mid){
            updataone(node*2,s,mid,v);
        }else{
            updataone(node*2+1,mid+1,e,v);
        }
        tree[node]=tree[node*2]+tree[node*2+1];
    }
}
int query(int node,int s,int e,int q,int p)
{
    if(s>p||e<q)
        return 0;
    if(q<=s&&e<=p)
        return tree[node];
    int mid=(s+e)/2;
    return query(node*2,s,mid,q,p)+query(node*2+1,mid+1,e,q,p);
}
int main()
{
    int sum,cot,mim;
    while(scanf("%d",&n)!=EOF)
    {
        sum=cot=0;
        mim=siz*4;
        memset(C,0,sizeof(C));//所有0-n-1都为0；
        build(1,0,n-1);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            C[a[i]]=1;//a[i]出现
            sum+=query(1,0,n-1,a[i],n-1);//求在a[i]前面且比a[i]大的数的个数
            updataone(1,0,n-1,a[i]);//a[i]已经出现，修改线段树，
        }
        mim=sum;
        for(int i=1;i<n;i++)
        {
            cot=sum+(n-2*a[i]-1);
            sum=cot;
            if(cot<mim) mim=cot;
        }
        cout<<mim<<endl;
    }
    return 0;
}