hdu 1394 Minimum Inversion Number （树状数组求逆序数）

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

输入n个数的序列，每次可以将第一个数放到最后面，求这个序列能形成的最小逆序数。

题解

先跑一个树状数组求逆序对，再利用相关的性质加加减减即可。

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int n, ans, minn, a[5010], t[5010], sum[5010];

int lowbit(int x) {
return x & (- x);
}

void modify(int pos) {
while(pos <= n) {
t[pos] += 1;
pos += lowbit(pos);
}
}

int query(int pos) {
int sum = 0;
while(pos) {
sum += t[pos];
pos -= lowbit(pos);
}
return sum;
}

int main() {
while(~ scanf("%d", &n)) {

4000
memset(t, 0, sizeof(t));
ans = 0;
for(int i = 1; i <= n; i ++)
scanf("%d", &a[i]);
for(int i = 1; i <=n; i ++) {
modify(a[i] + 1);
ans += i - query(a[i] + 1);
}
minn = ans;
for(int i = 1; i <= n; i ++) {
ans -= a[i];
ans += n - a[i] - 1;
minn = min(minn, ans);
}
printf("%d\n", minn);
}
return 0;
}