Minimum Inversion Number

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 84   Accepted Submission(s) : 62

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

101 3 6 9 0 8 5 7 4 2

 

Sample Output

16

 

Author

CHEN, Gaoli

 

Source

ZOJ Monthly, January 2003

 

#include <cstdio>#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1const int maxn = 55555;int sum[maxn << 2], g[5010];void PushUP(int rt){sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}void build(int l, int r, int rt)  // 向上建树{if (l == r){sum[rt] = 0;  // 初始化   都存0return;}int m = (l + r) >> 1;build(lson);build(rson);PushUP(rt);}void update(int p, int add, int l, int r, int rt)   //  P 处加上 add   单点更新{if (l == r){sum[rt] += add;return;}int m = (l + r) >> 1;if (p <= m) update(p, add, lson);else update(p, add, rson);PushUP(rt);}int query(int L, int R, int l, int r, int rt)   //   L  到  R 的和    线段求和{if (L <= l && r <= R){return sum[rt];}int m = (l + r) >> 1;int ret = 0;if (L <= m) ret += query(L, R, lson);if (R > m) ret += query(L, R, rson);return ret;}int main(){int n, s, min;while (scanf("%d", &n) != EOF){build(1, n, 1);s = 0;for (int i = 1; i <= n; i++){scanf("%d", &g[i]);s+=query(g[i]+1,n, 1, n, 1);    //  线段求和update(g[i]+1, 1, 1, n, 1);    //  单点更新}min = s;for (int i = 1; i < n; i++){s += n - 2 * g[i] - 1;   //   移动后的逆序书对if (s < min) min = s;}printf("%d\n", min);}return 0;}