ZOJ 1484 Minimum Inversion Number（线段树，数论）

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1484

Minimum Inversion Number

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

题意：

给出n个数（0~n-1，每个数仅出现一次）,问它长为n的循环序列中逆序对最少的数量。

分析：

如序列a1,a2,a3,a4,a5，它的逆序对数量s=sum(num(ak>ai,k<i));

序列变为a2,a3,a4,a5,a1，和上一个序列相比，变化分为两步，拿走a1和将a1 放入序列尾部;拿走a1,逆序对减少a1（ 整个序列是0~n-1且每个数字仅出现一次）个，加入序列尾部,逆序对增加n-1-a1个，这样就可以递推求解各序列的逆序对个数。

那么首个序列的逆序对如何求解呢？方法有很多，我采用的是线段树，维护[0,n)的区间和，每次先求出区间[ai+1,n)的和，即为ai之前比ai大的数字的个数,然后在ai位置上+1（初始化为0）,这样就能求解出首个序列的逆序对个数。

#include<cstdio>#include<iostream>#include<cstdlib>#include<algorithm>#include<ctime>#include<cctype>#include<cmath>#include<string>#include<cstring>#include<stack>#include<queue>#include<list>#include<vector>#include<map>#include<set>#define sqr(x) ((x)*(x))#define LL long long#define itn int#define INF 0x3f3f3f3f#define PI 3.1415926535897932384626#define eps 1e-10#define mm 5007using namespace std;int a[mm];itn _minv[mm<<2],_maxv[mm<<2];int _sumv[mm<<2];void update(itn p,int k,int l,int r){    if (r-l==1)    {        _sumv[k]=1;        return ;    }    int m=l+r>>1;    if (p<m)    update(p,k*2+1,l,m);  else    update(p,k*2+2,m,r);    _sumv[k]=_sumv[k*2+1]+_sumv[k*2+2];}int query(itn a,int b,int k,int l,int r){    if (b<=r || r<=a)   return 0;    if (a<=l && r<=b)   return _sumv[k];    return query(a,b,k*2+1,l,l+r>>1)+query(a,b,k*2+2,l+r>>1,r);}int main(){    #ifndef ONLINE_JUDGE        freopen("/home/fcbruce/文档/code/t","r",stdin);    #endif // ONLINE_JUDGE    int n;    while (~scanf("%d",&n))    {        for (int i=0;i<n;i++)            scanf("%d",a+i);        memset(_sumv,0,sizeof _sumv);        int s=0;        for (int i=0;i<n;i++)        {            s+=query(a[i]+1,n+1,0,0,n);            update(a[i],0,0,n);        }        itn ans=s;        int t=a[0];        for (int i=1;i<n;i++)        {            s=s+n-1-(t<<1);            t=a[i];            ans=min(ans,s);        }        printf("%d\n",ans);    }    return 0;}