线段树

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

题解1（暴力法）：直接暴力求解就行了；

#include <iostream>
#include <cstdio>
using namespace std;
#define inf 0x3f3f3f3f
int a[5500];
int main()
{
    int n,ans;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        int cnt=0;
        for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++)
                if(a[i]>a[j]) cnt++;
        ans=cnt;
        for(int i=0;i<n;i++)
        {
            cnt=cnt-a[i]+(n-1-a[i]);
            if(ans>cnt)  ans=cnt;
        }
        printf("%d\n",ans);
    }
    return 0;
}

题解2（线段树）：num表示 l，r 区间内已有的数字，每输入一个 a[i],先检查 a[i], n-1 区间内已有的数字，这就是 a[i] 之前比 a[i] 大的数字的数量，全部累加，即为所求的逆序; 再循环一遍更新最小逆序就行了;

#include <iostream>
#include <cstdio>
using namespace std;
#define inf 0x3f3f3f3f
const int maxn=5500;
int a[maxn];
struct seg
{
    int l,r,num;
}tree[maxn<<2];
void build(int p, int l, int r)
{
    tree[p].l = l;
    tree[p].r = r;
    tree[p].num = 0;
    if(l == r) return;
    int mid = (l + r) >> 1;
    build(p<<1, l, mid);
    build(p<<1|1, mid+1, r);
}
void add(int p, int x)
{
    if(tree[p].l == tree[p].r)
    {
        tree[p].num ++;
        return;
    }
    int mid = (tree[p].l + tree[p].r) >> 1;
    if(x<=mid) add(p<<1, x);
    else add(p<<1|1, x);
    tree[p].num = tree[p<<1].num + tree[p<<1|1].num;
}
int ask(int p, int l, int r)
{
    if(l == tree[p].l && tree[p].r == r) return tree[p].num;
    int mid = (tree[p].l + tree[p].r) >> 1;
    if(r <= mid) return ask(p<<1, l, r);
    else if(l > mid) return ask(p<<1|1, l, r);
    else return ask(p<<1, l, mid) + ask(p<<1|1, mid+1, r);
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        build(1,0,n-1);
        int sum=0;
        for(int i=0;i<n;i++)
        {
            sum+=ask(1,a[i],n-1);
            add(1,a[i]);
        }
        int ans=sum;
        for(int i=0;i<n;i++)
        {
            sum=sum-a[i]+n-1-a[i];
            if(ans>sum) ans=sum;
        }
        printf("%d\n",ans);
    }
    return 0;
}