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 5005using 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;}