Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6642 Accepted Submission(s): 4051
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.
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
#include<iostream> #include <cstring> #include <cstdio> #include <algorithm> using namespace std; #pragma warning(disable : 4996) const int MAXN = 5005; int tree[MAXN]; int num[MAXN]; int n; int lowbit(int x) { return x & (-x); } int getsum(int pos) { int sum = 0; while (pos > 0) { sum += tree[pos]; pos -= lowbit(pos); } return sum; } void update(int pos, int value) { while (pos <= n) { tree[pos] += value; pos += lowbit(pos); } } int main() { freopen("in.txt", "r", stdin); int sum, ans; while (scanf("%d", &n) != EOF) { sum = 0; memset(tree, 0, sizeof(tree)); for(int i = 1; i <= n; i++) { scanf("%d", &num[i]); update(num[i] + 1, 1); sum += getsum(n) - getsum(num[i] + 1); //sum += i - getsum(num[i] + 1); } //cout << sum << endl; ans = sum; for (int i = 1; i <= n; i++) { sum = sum - 2 * num[i] + n - 1; if(sum < ans) { ans = sum; } } printf("%d\n", ans); } return 0; }