hdu 1394 ------线段树入门题目-优快云博客

本文介绍了一种使用线段树解决最小逆序数问题的方法，包括核心算法原理、步骤详解及实例分析。

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Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13110 Accepted Submission(s): 8015

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

Author

CHEN, Gaoli

Source

ZOJ Monthly, January 2003

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刚学线段树，这道题也是上网看别人的代码，什么时候自己也能写出来啊。。。。。

题目大概意思是求最小逆序数。

注意：1、对于某一序列，其中的某一个数a[i]能构成多少个逆序，只须判断在a[i]+1~n的范围内找之前的数是否出现过的次数；
2、然后求出第一个序列的逆序数。
3、由第一个序列的逆序数可以推出它下一个序列的逆序数。有规律下一个序列的逆序数 sum = 上一个的sum - a[i] - 1 - a[i];如果得出呢比如说：
序列 3 6 9 0 8 5 7 4 2 1 把3移到后面，则它的逆序数会减少3个（0 2 1）但同时会增加 n - a[i] - 1个。对于1中如何求其出现的次数呢？具体见Update函数原理是，由开始往后更新，每输一个数就在其对应的位置标志其出现过。如输了 3 6 9 则 node[u].l = node[u].r = 3 6 9 的sum都为1 当输入0时，查找区间1～10内之前输入的有哪些数出现就是它的逆序数此时 3 6 9 都出现过了。

#include <iostream>
#include <stdio.h>
#define M 5005
using namespace std;
struct Tree
{
    int l,r,sum;
}tree[M*4];
int a[M];
void build(int pos,int l,int r)
{
    tree[pos].l=l;
    tree[pos].r=r;
    tree[pos].sum=0;
    if(tree[pos].l==tree[pos].r)
    {
        return;
    }
    int mid=(tree[pos].l+tree[pos].r)/2;
    build(pos*2,l,mid);
    build(pos*2+1,mid+1,r);
}
void update(int pos,int val)
{
    if(tree[pos].l==tree[pos].r)
    {
        tree[pos].sum++;
        return;
    }
    if(val<=tree[pos*2].r)
    update(pos*2,val);
    else
    update(pos*2+1,val);
    tree[pos].sum=tree[pos*2].sum+tree[pos*2+1].sum;
}
int query(int pos,int val)
{
    if(val<=tree[pos].l)
    return tree[pos].sum;
    if(val<=tree[pos*2].r)
    return query(pos*2,val)+query(pos*2+1,val);
    else return query(pos*2+1,val);
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        build(1,0,n);
        int sum=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum+=query(1,a[i]+1);
            update(1,a[i]);
        }
        int ans=sum;
        for(int i=0;i<n;i++)
        {
            sum+=n-a[i]-a[i]-1;
            ans=min(ans,sum);
        }
        printf("%d\n",ans);
    }
    return 0;
}