HDU 1394：Minimum Inversion Number（线段树区间求和单点更新）

http://acm.hdu.edu.cn/showproblem.php?pid=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

 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 5005
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
struct node
{
    int l,r,value;
}tree[N<<2];
int a[N];
/*
求移位后的最小逆序数
Update单点增减
Query区间求和
*/
void Build(int rt,int l,int r)
{
    tree[rt].l=l;
    tree[rt].r=r;
    tree[rt].value=0;
    if(l==r) return ;
    int m=(l+r)>>1;
    Build(rt<<1,l,m);
    Build(rt<<1|1,m+1,r);
}

void Update(int rt,int num)
{
    if(tree[rt].l==num&&tree[rt].r==num){
        tree[rt].value=1;
        return ;
    }
    int m=(tree[rt].l+tree[rt].r)>>1;
    if(num<=m) Update(rt<<1,num);
    else Update(rt<<1|1,num);
    tree[rt].value=tree[rt<<1].value+tree[rt<<1|1].value;
}

int Query(int rt,int l,int r)
{
/*
画个图就明白了，搜寻的区间为[l,r]，m代表当前节点的左儿子和右儿子的分支
如果左儿子有包含该区间的元素，即l<=m，那么就要继续递归搜寻左儿子
如果右儿子有包含该区间的元素，即m<r,那么就要继续递归搜寻右儿子
当当前的节点的区间被搜寻的区间[l,r]覆盖的话，就说明该段区间完全在[l,r]里面
那么返回该节点的值
*/
    if(l<=tree[rt].l&&tree[rt].r<=r){
        return tree[rt].value;
    }
    else{
        int m=(tree[rt].l+tree[rt].r)>>1;
        int s1=0,s2=0;
        if(l<=m) s1=Query(rt<<1,l,r);
        if(r>m) s2=Query(rt<<1|1,l,r);
        return s1+s2;
    }
}

int main()
{
    int n;
    while(~scanf("%d",&n)){
        Build(1,1,n);
        int sum=0;
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
            a[i]++;
            sum+=Query(1,a[i]+1,n);
            Update(1,a[i]);
        }
        int res = sum;
        for(int i=n-1;i>=0;i--){
            sum=sum-(n-a[i])+(a[i]-1);
            if(sum<res) res=sum;
        }
        printf("%d
",res);
    }
    return 0;
}