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