Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 19405    Accepted Submission(s): 11684 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 题意：求一个数组所有循环同构体中，逆序数最小的值。

思路：先用线段树求出原始序列的逆序数，后面的就可以推导出来了。

# include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # define MAXN 5000 # define lson l, m, id<<1 # define rson m+1, r, id<<1|1 using namespace std; int sum[MAXN<<2], a[MAXN+1]; void init(int l, int r, int id) { if(l == r) return; int m = (l+r)>>1; init(lson); init(rson); } void update(int p, int l, int r, int id) { if(l == r) { ++sum[id]; return; } int m = (l+r)>>1; if(p <= m) update(p, lson); else update(p, rson); sum[id] = sum[id<<1] + sum[id<<1|1]; } int query(int L, int R, int l, int r, int id) { if(L <= l && R >= r) return sum[id]; int m = (l+r)>>1; int ans = 0; if(L <= m) ans += query(L, R, lson); if(R > m) ans += query(L, R, rson); return ans; } int main() { int n, total; while(~scanf("%d",&n)) { total = 0; memset(sum, 0, sizeof(sum)); init(0, n-1, 1); for(int i=0; i<n; ++i) { scanf("%d",&a[i]); total += query(a[i], n-1, 0, n-1, 1); update(a[i], 0, n-1, 1); } int tmp = total; for(int i=0; i<n; ++i) { total += n - a[i] - a[i] - 1; tmp = min(tmp, total); } printf("%d\n",tmp); } return 0; }