Hdu 1394 Minimum Inversion Number 树状数组求逆序对

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

---

题目大意

给你一个0到n-1的排列，计算逆序对个数，然后再把第一个数移到最后面去，计算逆序对个数，然后再移新的第一个数到最后面去，计算逆序对个数，移完一圈之后，问逆序对个数最少的一次，逆序对个数是多少？

---

显然这道题可以用树状数组很快地处理出来，可以直接用树状数组求逆序对，然后再通过第一个逆序对去推后面的逆序对个数，观察一下可以发现，原来的第一个数被移到最后面去了的时候，其实新的贡献就是加上前面有多少大于他的数，然后再减去之前的贡献，也就是他在第一位的时候后面有多少个数比他小，然后更新答案，进行比较

---

#include<cstdio>#include<iostream>#include<cstring>using namespace std;const int MAXN =50000;int C[MAXN],a[MAXN],n,final;int lowbit(int x){return x&(-x);}void modify(int x,int val){    while(x<=n){        C[x]+=val;        x+=lowbit(x);    }} int query(int x){    int ass=0;    while(x){        ass+=C[x];        x-=lowbit(x);     }    return ass;}void init(){    memset(C,0,sizeof(C));    final=0;} int main(){    while(~scanf("%d",&n)){        init();        for(register int i=1;i<=n;i++){            scanf("%d",&a[i]);            modify(a[i]+1,1);            final+=i-query(a[i]+1);        }        int temp=final;        for(register int i=1;i<=n;i++){            temp+=(n-a[i]-1)-a[i];            final=min(final,temp);        }        printf("%d\n",final);    }    return 0;}