HDU1394 Minimum Inversion Number

/*
求逆序数
Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7097    Accepted Submission(s): 4333


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
 

Author
CHEN, Gaoli
 

Source
ZOJ Monthly, January 2003 
 

Recommend
Ignatius.L

*/
/*暴力解决*/
#include <iostream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
#define maxn  5100
#define Max 100000000
int main()
{
    int sum[maxn],ans,i,j,n,num;
    while(~scanf("%d",&n))
    {
        ans=Max;
        num=0;
        for(i=1;i<=n;i++)
        scanf("%d",&sum[i]);
        for(i=1; i<=n; i++)
            for(j=i+1; j<=n; j++)
            {
                if(sum[i]>sum[j])
                    num++;
            }
        for(i=1;i<=n;i++)
        {
            num=num+(n-sum[i])-(sum[i]+1);
            ans = min(ans,num);
        }
        printf("%d
",ans);
    }
    return 0;
}

/*用线段树做*/
#include <iostream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
//#define maxn 50100
const int maxn = 222222;
#define lson l, m , rt<<1
#define rson m+1 , r ,rt<<1|1
int  sum[maxn<<2];
void PushUP(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void cre(int l,int r,int rt)
{
    sum[rt]=0;
    if(l==r)
    {
        return ;
    }
    int m = (l+r)>>1;
    cre(lson);
    cre(rson);
    PushUP(rt);
}
void update(int p, int l,int r,int rt)
{
    if(l==r)
    {
        sum[rt]++;
        return ;
    }
    int m = (l + r)>>1;
    if(p<=m) update(p,lson);
    else update(p,rson);
    PushUP(rt);
    // 和方法2对照发现两个递归有点区别，法2是先把根节点的和更新，而
    //法3是先更新叶子节点，然后反过来更新根节点
}
int qur(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        return sum[rt];
    }
    //大神是分数组的区间，而我们是分要求的区间，所以他是小于等于，真妙
    int m = (l + r)>>1;
    int ret = 0;
    if(L<=m)
        ret+=qur(L,R,lson);
    if(R>m)
        ret+=qur(L,R,rson);
    return ret;
}
int a[maxn];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        cre(0,n-1,1);
        int sum=0;
        for(int i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
            sum+=qur(a[i],n-1,0,n-1,1);
            update(a[i],0,n-1,1);
        }
        int ret =sum;
        for(int i=0; i<n; i++)
        {
            sum+=n-(a[i]+1)-a[i];
            ret=min(ret,sum);
        }
        printf("%d
",ret);
    }
    return 0;
}