HDU Problem 1394 Minimum Inversion Number【树状数组】

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 18277    Accepted Submission(s): 11093


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

 

Author

CHEN, Gaoli

 

Source

ZOJ Monthly, January 2003

 

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一开始没有看到数组是连续的。。。。

先求出最原始的逆序数和ans

如果数组是连续的那就好办了，对于第一个元素，有a【0】 - 1 个元素是小于它的， 而有n - a【0】‘个元素是大于它的。所以每次转换就变成了

ans = ans - a【0】+ 1 + n +  - a【0】；

#include <cstdio>
#include <set>
#include <map>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#define space " "
using namespace std;
const int MAXN = 5000 + 10;
const int INF = 0x3f3f3f3f;
int n, ar[MAXN], b[MAXN], num[MAXN];
int sum(int x) {
int s = 0;
while (x >= 1) {
s += num[x];
x -= x&-x;
}
return s;
}
void add(int x, int y) {
while (x <= n) {
num[x] += y;
x += x&-x;
}
}
int main() {
while (scanf("%d", &n) != EOF) {
for (int i = 1; i <= n; i++) {
scanf("%d", &ar[i]); ar[i]++;
}
int ans = 0;
memset(num, 0, sizeof(num));
for (int i = 1; i <= n; i++) {
add(ar[i], 1);
ans += i - sum(ar[i]);
}
int minn = ans;
for (int i = 1; i <= n; i++) {
ans += n - ar[i] - ar[i] + 1;
minn = min(minn, ans);
}
printf("%d\n", minn);
}
return 0;
}