HDU 1394 Minimum Inversion Number （暴力+線段樹）

Mr_Treeeee發表於2020-04-06

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20570    Accepted Submission(s): 12335


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 



 





題意：


重複一次操作可以把最前面的樹放到最後面。求最小的逆序總數，逆序數是i<j且Ai>Bi。





POINT:


因為給你N後數是確定的，是0-(n-1)，所以只要知道最前面的數的值，就可以知道把它移到後面後逆序總數的變化。


比第一個小的數量必為Ai， 大的必為n-Ai-1。這是這題的關鍵。


下面則是要先算出初始狀態時的逆序總數，再根據它一次一次移動求出每個的逆序總數，在和ans比較。





這裡有兩種方法求初始狀態時的逆序總數。


法1:暴力，每輸入一個數，去找前面有幾個比他大的，記錄答案。顯然這個非常簡單。


法2:利用線段樹，此時的線段樹的區間直接代表數值，存的則是這些數出現了幾次。


即num[x]代表值在l和r的數出現了幾次。也是每輸入一個數，存下這個數，更新區間，再去詢問一次Ai+1到n+1的數出現了幾次。





法1:顯然比線段樹短很多，但是也慢很多。


#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;
const int N = 5000+5;
int a[N];
int sum[N];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(sum,0,sizeof sum);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            for(int j=1;j<i;j++)
            {
                if(a[j]>a[i])
                {
                    sum[i]++;
                }
            }
        }
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            ans+=sum[i];
        }
        int pre=ans;
        for(int i=1;i<=n;i++)
        {
            int now=pre;
            int xiao=a[i];
            int da=n-a[i]-1;//不包括自己
            pre=now+da-xiao;
            ans=min(ans,pre);
        }
        printf("%d\n",ans);
    }

    


法2:


#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;
const int N = 5000*4;
int a[N/4];
int num[N];
void add(int x,int l,int r,int now)
{
    if(l==r&&r==now) num[x]++;
    else
    {
        int mid=(l+r)>>1;
        if(now<=mid) add(x*2,l,mid,now);
        if(mid<now) add(2*x+1,mid+1,r,now);
        num[x]=num[2*x+1]+num[x*2];
    }
}
int query(int x,int l,int r,int ll,int rr)
{
    int ans=0;
    if(ll<=l&&rr>=r) ans=num[x];
    else
    {
        int mid=(l+r)>>1;
        if(ll<=mid) ans+=query(x*2,l,mid,ll,rr);
        if(mid<rr) ans+=query(2*x+1,mid+1,r,ll,rr);
    }
    return ans;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(num,0,sizeof num);
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            add(1,0,n-1,a[i]);
            ans+=query(1,0,n-1,a[i]+1,n-1);
        }
        int pre=ans;
        for(int i=1;i<=n;i++)
        {
            int now=pre;
            now=pre-a[i]+n-a[i]-1;
            ans=min(now,ans);
            pre=now;
        }
        printf("%d\n",ans);
    }
    
    
}



                                    




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