HDU 1394 线段树

Minimum Inversion Number

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

题目大意：
给一个n个数的序列a1, a2, ..., an ，这些数的范围是0～n-1， 可以把前面m个数移动到后面去，形成新序列：
a1, a2, ..., an-1, an
a2, a3, ..., an, a1

a3, a4, ..., an, a1, a2

...
an, a1, a2, ..., an-1
求这些序列中，逆序数最少的是多少？

求出a1, a2, ..., an-1, an的逆序数之后，就可以递推求出其他序列的逆序数。 假设要把a1移动到an之后，那么我们把这个过程拆分成两步：
1.  把a1去除掉。通过观察可以发现，（a1-1）是0～n-1中比a1小的数字的个数，由于a1在序列的第一个所以a1之后共有(a1-1)个比a1小，所以形成了（a1-1）对逆序数，当去除掉a1时，原序列的逆序数总数也就减少了（a1-1）个逆序数。 
2. 把a1加到an之后。0～n-1中，比a1大的数共有（n-a1）个数，由于a1现在在最后一个，也就是它前面共有（n-a1）个数比它大，即增加了（n-a1）对逆序数。
综合1，2两步， 设原序列逆序数为sum, 当把原序列第一个移动到最后位置时，逆序数变为：sum = sum-(ai-1)+(n-ai);

（方法是度娘的...）

代码是自己的..

#include <iostream>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const int maxn=111111;int sum[maxn<<2],x;int s[maxn<<2];void pushup(int rt){sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void build(int l,int r,int rt){sum[rt]=0;if(l==r)return;int m=(l+r)>>1;build(lson);build(rson);}void update(int l,int r,int rt,int s){if(l==r){sum[rt]=1;return;}int m=(l+r)>>1;if(s<=m)update(lson,s);else if(s>m)update(rson,s);pushup(rt);}void query(int l,int r,int rt,int L,int R){if(L<=l&&R>=r){x+=sum[rt];return;}int m=(l+r)>>1;if(L<=m)query(lson,L,R);if(R>m)query(rson,L,R);}int main(){int n,i,j,min;while(~scanf("%d",&n)){build(0,n-1,1);x=0;for(i=0;i<n;i++){scanf("%d",&s[i]);query(0,n-1,1,s[i],n-1);update(0,n-1,1,s[i]);}min=x;for(i=0;i<n;i++){x=x-s[i]+(n-1)-s[i];if(min>x)min=x;}printf("%d\n",min);}return 0;}