hdu 1394 逆序对（nlgn+o(n) ）

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20404 Accepted Submission(s): 12231

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

思路:先求出原序列的逆序对和，在递推求变换位置的和，取最小值，用线段树叶子维护0~n-1的值

#include <cstdio>#include <iostream>#include <algorithm>using namespace std;#define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1const int maxn=5000+100;int sum[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);//no need pushup all 0}void update(int p,int l,int r,int rt) {    if(l==r) {        sum[rt]=1;        return ;    }    int m=(l+r)>>1;    if(p<=m) update(p,lson);    else update(p,rson);    pushup(rt);}int query(int L,int R,int l,int r,int rt) {    if(L<=l&&r<=R) {        return sum[rt];    }    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 a[maxn];int main() {//  freopen("input.txt","r",stdin);    int n;    while(scanf("%d",&n)!=EOF) {        int ans=0;        build(0,n-1,1);        for(int i=0;i<n;i++) {            scanf("%d",&a[i]);            ans+=query(a[i],n-1,0,n-1,1);            update(a[i],0,n-1,1);        }        int ret=ans;        for(int i=0;i<n;i++) {            ans+=n-a[i]-a[i]-1;            ret=min(ret,ans);        }        printf("%d\n",ret);    }    return 0;}