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：

用线段树求逆序对。只需要求出第一组逆序对就能推出移项之后的逆序对。假设一组序列有n个元素，元素包含0 ~ n - 1，未移项时逆序对有m组，第一个元素为a，那么当第一个元素移到最后时，比第一个元素小的元素都失去了这组逆序对，所以少了c组逆序对，而比第一个元素大的元素都多了一组逆序对，所以多了 n - 1 - c。因此移项一次项后的逆序对为m - c + (n - 1 - c)组。通过这个公式不断递推即可求得每次移项后的逆序对个数，从而求出最小的个数。

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){    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){    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){    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(0, n - 1, 1);        for(int i = 0; i < n; i++)        {            scanf("%d", &a[i]);            sum += query(a[i] + 1, n - 1, 0, n - 1, 1);            update(a[i], 0, n - 1, 1);        }        ans = sum;        for(int i = 0; i < n - 1; i++)        {            sum += n - 1 - 2 * a[i];            ans = min(ans, sum);        }        printf("%d\n", ans);    }    return 0;}