Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 19452    Accepted Submission(s): 11701 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

大致题意：

对于给出的数组，每次把第一个元素挪到最后，求此过程中最小的逆序数。

大体思路：

可以用归并排序求。这里说用线段树的解法。

处理初始数组：对于每一个数，查找线段树中已有的比他大的数的数量。之后把这个数加入线段树。

每一次移动：假设这次移动的是m，则他后面比m小的有 m-1 个，比m大的有 n-m个，所以新的逆序数等于 原来的逆序数 - （m-1） + n - m.

#include<cstdio> #include<cstring> #include<iostream> using namespace std; #define _max 5010 int Seg [_max<<2]; int A [_max] ; int n ; int Query (int p, int l, int r, int x, int y) { if(x<=l && y>=r) return Seg[p]; int mid = (l+r)>>1,ans = 0; if(x<=mid) ans += Query(2*p, l, mid, x, y); if(y> mid) ans += Query(2*p+1, mid+1, r, x, y); return ans; } void Update (int p, int l, int r, int x) { if(l==r){ Seg[p]++; return; } int mid = (l+r)>>1; if(x<=mid) Update(2*p, l, mid, x); else Update(2*p+1, mid+1, r, x); Seg[p] = Seg[2*p] + Seg[2*p+1]; } int main () { //freopen("in.txt","r",stdin); while(scanf("%d",&n) != EOF){ memset(Seg,0,sizeof(Seg)); for( int i=1; i<=n; i++){ scanf("%d",A+i); A[i]++; } int ans = 0 ; for(int i=1; i<=n; i++){ ans += Query(1, 1, n, A[i], n); Update(1, 1, n, A[i]); } int Min = ans; for(int i=1; i<=n; i++){ ans = ans - A[i] + 1 + n - A[i] ; if(ans<Min) Min = ans ; } printf("%d\n",Min); } return 0 ; }