HDU-1394 Minimum Inversion Number（线段树求逆序数）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 15675 Accepted Submission(s): 9569

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

这道题目有两个地方可以留意一下。

首先求一个数列的逆序数

首先将这个数列排序，然后从最后一个数，边询问边更新

第二，根据题意，将第一个数放到最后一个，形成的新数列应该怎么求逆序数。其实可以递推，和没有变化前的数列比较逆序数增加了(n-1-a[i])-a[i]

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>

using namespace std;
#define MAX 5000
int segTree[MAX*4+5];
int n;
int a[MAX];
int b[MAX];
int c[MAX];
void PushUp(int node)
{
    segTree[node]=segTree[node<<1]+segTree[node<<1|1];
}
void build(int node,int begin,int end)
{
    if(begin==end)
    {
        segTree[node]=0;
        return;
    }
    int m=(begin+end)>>1;
    build(node<<1,begin,m);
    build(node<<1|1,m+1,end);
    PushUp(node);
}
void Update(int node,int begin,int end,int ind,int num)
{
    if(begin==end)
    {
        segTree[node]+=num;
        return;
    }
    int m=(begin+end)>>1;
    if(ind<=m)
        Update(node<<1,begin,m,ind,num);
    else
        Update(node<<1|1,m+1,end,ind,num);
    PushUp(node);
}
int Query(int node,int begin,int end,int left,int right)
{
    if(left<=begin&&end<=right)
        return segTree[node];
    int ret;
    int m=(begin+end)>>1;
    ret=0;
    if(left<=m)
        ret+=Query(node<<1,begin,m,left,right);
    if(right>m)
        ret+=Query(node<<1|1,m+1,end,left,right);
    return ret;
}
int cmp(int x,int y)
{
    return x<y;
}
int main()
{
    int sum;
    while(scanf("%d",&n)!=EOF)
    {
        sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            c[i]=a[i];
            b[a[i]]=i;
        }
        build(1,1,n);
        sort(a+1,a+n+1,cmp);
        for(int i=n;i>=1;i--)
        {
            sum+=Query(1,1,n,1,b[a[i]]);
            Update(1,1,n,b[a[i]],1);
        }
        int num;
        num=sum;
        int ans;
        ans=sum;
        for(int i=1;i<n;i++)
        {
            num+=-c[i]+(n-1-c[i]);
            if(ans>num)
                ans=num;
        }
        printf("%d\n",ans);

    }
    return 0;