[HDU] 1394 Minimum Inversion Number [线段树求逆序数]

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11788    Accepted Submission(s): 7235

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

 

Recommend

Ignatius.L

 

题解：这题的维护信息为每个数是否已经出现。每次输入后，都从该点的值到n-1进行查询，每次发现出现了一个数，由于是从该数的后面开始找的，这个数肯定是比该数大的。那就是一对逆序数，然后逆序数+1.最后求完所有的逆序数之后，剩下的就可以递推出来了。因为假如目前的第一个数是x，那当把他放到最后面的时候，少的逆序数是本来后面比他小的数的个数。多的逆序数就是放到后面后前面比他大的数的个数。因为所有数都是从0到n-1.所以比他小的数就是x，比他大的数就是n-1-x。这样的话每次的逆序数都可以用O(1)的时间计算出来。然后找最小的时候就可以了。

 

 #include<cstdio>
 #include<algorithm>

 using namespace std;

 #define  lson  l , m , rt << 1
 #define  rson  m + 1 , r , rt << 1 | 1

 const int maxn =  + ;
 int sum[maxn<<];

 void PushUp(int rt)
 {
     sum[rt]=sum[rt<<]+sum[rt<<|];
 }

 void build(int l,int r,int rt)
 {
     int m;

     sum[rt]=;
     if(l==r) {
         return;
     }
     m=(l+r)>>;
     build(lson);
     build(rson);
 }

 int query(int L,int R, int l, int r,int rt)
 {
     int m,ret;

     if(L<=l && r<=R) {
         return sum[rt];
     }

     m=(l+r) >> ;
     ret=;
     if(L<=m) ret+=query(L,R,lson);
     if(R>m) ret+=query(L,R,rson);

     return ret;
 } 

 void Updata(int p,int l,int r,int rt)
 {
     int m;
     if (l==r) {
         sum[rt]++;
         return ;
     }
     m=(l+r)>>;
     if(p<=m) Updata(p,lson);
     else Updata(p,rson);

     PushUp(rt);
 }

 int main()
 {
     int n,sum,x[maxn];

     while(~scanf("%d",&n)) {
         sum=;
         build(,n-,);

         for(int i = ;i < n;i++) {
             scanf("%d",&x[i]);
             sum+=query(x[i],n-,,n-,);
             Updata(x[i],,n-,);
         }
         int ret=sum;
         for(int i=;i<n;i++) {
             sum+=n-x[i]-x[i]-;
             ret=min(ret,sum);
         }

         printf("%d\n",ret);
     }

     return ;
 }