hdu1394 Minimum Inversion Number（求逆序对）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3466    Accepted Submission(s): 2113

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

 

题意：共有n-1个序列求最小的逆序对；

方法一：归并 O(n^2);

#include<iostream>
#define N 5010
using namespace std;

int A[N],B[N],T[N],ans;

void merge(int s1,int e1,int s2,int e2)
{
    int k,i,j;
    k=i=s1;j=s2;
    while(i<=e1&&j<=e2)
    {
        if(A[i]<=A[j])
        {
            T[k]=A[i];
            k++;i++;
        }
        else
        {
            T[k]=A[j];
            ans+=(e1-i+1);//i后面的肯定都大于A[j]
            k++;j++;    
        }
    }
    while(i<=e1)
    {
        T[k++]=A[i++];
    }
    while(j<=e2)
    {
        T[k++]=A[j++];
    }
    for(i=s1;i<=e2;i++) A[i]=T[i];
}

void mergesort(int s,int e)
{
    if(s<e)
    {
        int m=(s+e)>>1;
        mergesort(s,m);
        mergesort(m+1,e);
        merge(s,m,m+1,e);
    }
}

int main()
{
    int n,i,j,t,t1,t2,tt,ttt;
    while(scanf("%d",&n)!=EOF)
    {
        ans=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&A[i]);
            B[i]=A[i];
        }
        mergesort(0,n-1);
        tt=ans;
        for(i=0;i<n-1;i++)
        {
            t1=t2=0;
            t=B[0];
            for(j=1;j<n;j++)
            {
                B[j-1]=B[j];
                if(t<B[j])t1++;
                else if(t>B[j]) t2++;
            }
            tt=tt+t1-t2;
            if(ans>tt) ans=tt;
            B[n-1]=t;
        }
        printf("%d\n",ans);
    }
    return 0;
}

 

方法二：线段树实现（这也是做这题的原因）（模板）

1、离散化：把起初的元素用 0..n-1 这些数来替代，如2 5 2 6 离散化就是0 0 1 2，不过这个题不要离散化，而已每个元素都是不一样的 = =！！

          这样的话你就很快知道比s[i]大的数有多少，比s[i]小的数有多少。

2、每段树中区间 [tree[r].left,tree[r].right] 中有多少个以前出现过。

3、每次输入一个key ，就查找( key,n-1 ]中有多少个数以前出现过，再把 key 插入树中。

4、假设t是当前逆序对的个数，把s[0]移到最后产生的新序列的逆序对为 t+(n-1-s[0])-s[0];循环n-1次就能找到最小值了。

#include<iostream>
#define N 5010
using namespace std;

struct segtree
{
    int left,right,sum;
}tree[N<<2];
int s[N];

void build(int r,int a,int b)
{
    tree[r].left=a;
    tree[r].right=b;
    tree[r].sum=0;
    if(a<b)
    {
        int m=(a+b)>>1;
        build(r<<1,a,m);
        build(r<<1|1,m+1,b);
    }
}

/*
没考虑和a一样大的,也不存在
在插入a之前有多少(a,b]之间的数已经插入了，都是逆序对
*/
int query(int r,int a,int b)
{
    if(tree[r].left==a&&tree[r].right==b)
        return tree[r].sum;
    int m=(tree[r].left+tree[r].right)>>1;
    if(b<=m) 
        return query(r<<1,a,b);
    else if(a>m)
        return query(r<<1|1,a,b);
    else 
        return query(r<<1,a,m)+query(r<<1|1,m+1,b);
}

void update(int r,int key)
{
    tree[r].sum++;
    if(tree[r].left==tree[r].right) return ;
    int m;
    m=(tree[r].left+tree[r].right)>>1;
    if(key<=m) 
        update(r<<1,key);
    else
        update(r<<1|1,key);
}

int main()
{
    int n,i,t,ans;
    while(cin>>n)
    {
        ans=0;
        build(1,0,n-1);
        for(i=0;i<n;i++)
        {
            cin>>s[i];
            ans+=query(1,s[i],n-1);
            update(1,s[i]);
        }
        t=ans;
        for(i=0;i<n-1;i++)
        {
            t=t+(n-1-s[i])-s[i];
            if(ans>t)
                ans=t;
        }
        cout<<ans<<endl;
    }
    return 0;
}