HDU 1394 Minimum Inversion Number（归并排序 线段树 逆序数）

http://acm.hdu.edu.cn/showproblem.php?pid=1394

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：

有n个从0到n 的数 按照给定次序排列 每次只能将最前的一个数移到末尾 算出每种情况的逆序数 并求出逆序数最小值

思路：

归并排序或线段树

首先 我们把当前情况下的逆序数求出

然后循环n次 每次最前一个往最后移 则 ans=ans+ (n-1+a[i])-a[i]

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;
int num[5000+10],a[5000+10],t[5000+10];
//int vis[5000+10];
int ans;

void Merge(int x,int m,int y)
{
    int p=x,q=m+1,i=x;
    while(p<=m||q<=y)
    {
        if(q>y||(p<=m&&a[p]<=a[q])) t[i++]=a[p++];
        else
        {
            t[i++]=a[q++];
            ans+=m-p+1;
        }
    }
    for(i=x;i<=y;i++) a[i]=t[i];
}
void mergesort(int x,int y)
{
   if(x!=y)
   {
       int mid=(x+y)/2;
       mergesort(x,mid);
       mergesort(mid+1,y);
       Merge(x,mid,y);
   }
}
int main()
{
    int n,i,j,minn;
    while(scanf("%d",&n)!=EOF)
    {
      ans=0;
      for(i=1;i<=n;i++)
        {scanf("%d",&a[i]); num[i]=a[i];}
      mergesort(1,n);
      minn=ans;
      for(i=1;i<n;i++)
      {
          ans=ans+n-1-2*num[i];
          if(ans<minn) minn=ans;
      }
      cout<<minn<<endl;
    }
    return 0;
}

　　

　　

线段树的版本 

对于初始序列

每次加入一个数就对其标记 并且查询是否有比他大的数已经被加入 从而求出逆序数

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#define mem(a,b) memset(a,b,sizeof(a))
#define ll __int64
#define MAXN 1000
#define INF 0x7ffffff
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
int sum[50000];
int a[5000+10];
void pushup(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt)
{
    if(l==r)
    {
        sum[rt]=0; return ;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    pushup(rt);
}
int getnum(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {

        return sum[rt];
    }
    int ret=0;
    int m=(l+r)>>1;
    if(L<=m) ret+=getnum(L,R,lson);
    if(R>m)  ret+=getnum(L,R,rson);
    return ret;
}
void update(int num,int l,int r,int rt)
{   
    if(l==r&&l==num)
    {       
        sum[rt]=1;
        return ;
    }    
    int m=(l+r)>>1;   
    if(num<=m) update(num,lson);
    else       update(num,rson);
    pushup(rt);
}
int main()
{
    int n;
    int i,j;
    int ans,minn;
    while(scanf("%d",&n)!=EOF)
    {
        ans=0;
        build(0,n-1,1);
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            ans+=getnum(a[i],n-1,0,n-1,1);          
            update(a[i],0,n-1,1);
        }
        minn=ans;       
        for(i=0;i<n-1;i++)
        {
            ans=ans-a[i]*2+n-1;
            if(minn>ans) minn=ans;
        }
        cout<<minn<<endl;
    }
    return 0;
}