hdu1394 求最小逆序数

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.

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.

没有看懂为什么要用线段树做tvt

因为数据给得很凑巧（0..n-1），数据也小，就暴力了。

#include <iostream>

using namespace
std;

int num[5050];

int main()

{

    int n;

    while (cin >> n) {

        

        for (int i =
0; i < n; i ++) {

            cin >>
num[i];

        }

        int sum =
0;

        for (int i =
0; i < n; i ++) {

            for (int j = i +
1; j < n; j ++) {

                if (num[j] <
num[i]) {

                    sum ++;

                }

            }

        }

        int mini = sum;

        for (int i =
0; i < n; i ++) {

            sum = sum - num[i] + n -
1 -
num[i];//当把i从第一个放到最后一个时，逆序数增加(在num[1..n
- 1]中)比i大的数的个数，减少比i小的数的个数

            mini = min(sum,mini);//因为数从0..n - 1，所以比i大的数有n
- 1 - i，比i小的数有i个(0..i
- 1),

        }

        cout << mini <<
endl;

    }

   

    return
0;

}