Minimum Inversion Number(hdu1394(线段数or暴力))

/*
http://acm.hdu.edu.cn/showproblem.php?pid=1394
Minimum Inversion Number


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7678    Accepted Submission(s): 4708




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
解析：
题意：
给出由0到n-1个数组成的数组,按照题目的轮换要求，求最少的逆序对
思路：
1.首先要求出初始串有多少个逆序对，这里有两种方法：
一：暴力求：由于数据不大此法可行
二;线段数，按照数组的顺序，进行逐次查询和更新，注意：无论是查询还是更新的区间都是为数而不是序号
2.利用规律，更新最小值。
这里解释一下sum=sum+n-a[i]-a[i]-1;
由于数组的数值只有0到n-1且不重复。则对于a[i]来说一定有a[i]个数比它小，n-a[i]-1个数比它大；
假设a[i]的前一状态是在数组首部而此时被放到数组尾部时，逆序对必然减少a[i]；增加n-a[i]-1.

/ 方法一暴力法：250MS 260K482 BC++#include<stdio.h>#include<string.h>#include<math.h>#include <iostream>using namespace std;const int maxn=5000+10;int a[maxn];int main(){int n,i,j; while(scanf("%d",&n)!=EOF) { for(i=1;i<=n;i++) { scanf("%d",&a[i]); } int s=0,ans; for(i=1;i<=n;i++)  { for(j=1;j<i;j++)   {   if(a[j]>a[i])    s++;   }}ans=s;for(i=1;i<=n-1;i++){s=s+n-1-2*a[i];if(s<ans)ans=s;}printf("%d\n",ans); }return 0;}// 方法二线段树：46MS312K1467 B#include<stdio.h>#include<string.h>#include<math.h>#include <iostream>using namespace std;const int maxn=5000+10;int a[maxn];struct tree{int l;int r;int num;}tr[4*maxn];void build(int k,int l,int r){tr[k].l=l;tr[k].r=r;tr[k].num=0;if(l==r){return;}int mid=(tr[k].l+tr[k].r)/2;build(k<<1,l,mid);build(k<<1|1,mid+1,r);//tr[k].num=tr[k<<1].num+tr[k<<1|1].num;}void update(int k,int l,int r){if(l==tr[k].l&&r==tr[k].r){tr[k].num++;return;}int mid=(tr[k].l+tr[k].r)/2;if(r<=mid)update(k<<1,l,r);else if(l>mid)update(k<<1|1,l,r);else{update(k<<1,l,mid);update(k<<1|1,mid+1,r);}tr[k].num=tr[k<<1].num+tr[k<<1|1].num;}int query(int k,int l,int r){if(l==tr[k].l&&r==tr[k].r){return tr[k].num;}int mid=(tr[k].l+tr[k].r)/2;if(r<=mid) return query(k<<1,l,r);else if(l>mid) return query(k<<1|1,l,r);else{  returnquery(k<<1,l,mid)+query(k<<1|1,mid+1,r);}}int main(){int n,i;while(scanf("%d",&n)!=EOF){int ans,sum;sum=0;for(i=1;i<=n;i++){scanf("%d",&a[i]);}build(0,0,n);for(i=1;i<=n;i++){if(a[i]<n)sum+=query(1,a[i]+1,n);//统计将比a[i]大的个数update(1,a[i],a[i]);//把a[i]插入树中并更新，这里是按照数列的输入顺序进行更新的；}ans=sum;for(i=1;i<=n-1;i++){sum=sum+n+1-2*a[i];//因为前if(sum<ans)ans=sum;}printf("%d\n",ans);}return 0;}