Minimum Inversion Number（线段树）

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

 101 3 6 9 0 8 5 7 4 2 

 

Sample Output

 16 

解题思路

之前做了两题都是完全套的模板，这题用来求逆序对一下转不过来了，于是看看骆大神的代码，原来如此！

题中已经有对逆序对的简单介绍，但还是在此贴下逆序数的概念：

在一个排列中，如果一对数的前后位置与大小顺序相反，即前面的数大于后面的数，那么它们就称为一个逆序。一个排列中逆序的总数就称为这个排列的逆序数。

逆序数为偶数的排列称为偶排列；逆序数为奇数的排列称为奇排列； 如2431中，21，43，41，31是逆序，逆序数是4，为偶排列。

也是就说，对于n个不同的元素，先规定各元素之间有一个标准次序（例如n个 不同的自然数，可规定从小到大为标准次序），

于是在这n个元素的任一排列中，当某两个元素的先后次序与标准次序不同时，就说有1个逆序。一个排列中所有逆序总数叫做这个排列的逆序数。

*那么如何用线段树做呢，build时都初始化为0，每次输入后先query，找出前面已经输入的数中比自己大的数，怎么找的呢，每次会对输入的数update，值为1；

AC代码

#include <iostream>#include <cstdio>using namespace std;#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1const int maxn = 5005;int sum[maxn << 2];void PushUp(int rt){    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}void build(int l, int r, int rt = 1){    sum[rt] = 0;    if(l == r)        return;    int m = (l + r) >> 1;    build(lson);    build(rson);    PushUp(rt);}void update(int p, int l, int r, int rt = 1){    if(l == r)    {        sum[rt]++;        return;    }    int m = (l + r) >> 1;    if(p <= m)        update(p, lson);    else        update(p, rson);    PushUp(rt);}int query(int L, int R, int l, int r, int rt = 1){    if(L <= l && r <= R)        return sum[rt];    int m  = (l + r) >> 1;    int ret = 0;    if(L <= m)        ret += query(L, R, lson);    if(m < R)        ret += query(L, R, rson);    return ret;}int main(){    int n, a[maxn], sum, ans;    while(scanf("%d", &n) != EOF)    {        sum = 0;        build(1, n);        for(int i = 1; i <= n; i++)        {            scanf("%d", &a[i]);            sum += query(a[i] + 1, n, 1, n);            update(a[i], 1, n);        }        ans = sum;        for(int i = 1; i <= n; i++)        {            sum += n - 1 - 2 * a[i];            ans = min(ans, sum);        }        printf("%d\n", ans);    }    return 0;}