Minimum Inversion Number（树状数组求逆序数+找数学规律）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=1394

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12203    Accepted Submission(s): 7446

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

 

Author

CHEN, Gaoli

 

Source

ZOJ Monthly, January 2003

 

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AC  code：

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<algorithm>#include<queue>#define MAXN 5010using namespace std;int c[MAXN],n,num[MAXN];int lowbit(int p){return p&(-p);}void up(int p,int v){while(p<=n){c[p]+=v;p+=lowbit(p);}}int sum(int p){int s=0;while(p>0){s+=c[p];p-=lowbit(p);}return s;}int main(){int t,ans;while(~scanf("%d",&n)){ans=0;memset(c,0,sizeof(c));for(int i=1;i<=n;i++){scanf("%d",&num[i]);up(num[i]+1,1);//注意加1！！！并且必须先更新再求和ans+=i-sum(num[i]+1); }int tmp=ans;for(int i=1;i<=n-1;i++){tmp=tmp-num[i]+(n-1-num[i]);ans=min(tmp,ans);}printf("%d\n",ans);}return 0;}

附上第二种做法（用线段树）：

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#define MAXN 5001using namespace std;struct node{int l,r,sum;}tree[MAXN*4];int ans,n;int num[MAXN];void build(int rt,int s,int e){tree[rt].l=s;tree[rt].r=e;if(s==e){tree[rt].sum=0;return;}int mid=(s+e)/2;build(rt*2,s,mid);build(rt*2+1,mid+1,e);tree[rt].sum=tree[rt*2].sum+tree[rt*2+1].sum;}void update(int rt,int pos,int val){if(tree[rt].l==pos&&tree[rt].r==pos){tree[rt].sum+=val;return;}int mid=(tree[rt].l+tree[rt].r)/2;if(pos<=mid){update(rt*2,pos,val);}else{update(rt*2+1,pos,val);}tree[rt].sum=tree[rt*2].sum+tree[rt*2+1].sum;}void query(int rt,int ql,int qr){int mid=(tree[rt].l+tree[rt].r)/2;if(tree[rt].l==ql&&tree[rt].r==qr){ans+=tree[rt].sum;}else if(qr<=mid){query(rt*2,ql,qr);}else if(ql>mid){query(rt*2+1,ql,qr);}else{query(rt*2,ql,mid);query(rt*2+1,mid+1,qr);}}int main(){int i,ans1;while(~scanf("%d",&n)){build(1,0,n-1);ans1=0;for(i=0;i<n;i++){scanf("%d",&num[i]);ans=0;query(1,num[i],n-1);ans1+=ans;        update(0,num[i],1);}int t=ans1;for(int i=0;i<n-1;i++){ans1=ans1-num[i]+(n-1-num[i]);t=min(t,ans1);}    printf("%d\n",t);}return 0; } 

以下拓展内容参考自http://blog.csdn.net/wiking__acm/article/details/7920429：

拓展一下，如果要求正序数怎么办？很简单，无非是大小调一下

再问，如果要求满足i<j<k,且a[i]>a[j]>a[k]的数对总数怎么办？

可以从中间的这个数入手，统计a[i]>a[j]的对数m，以及a[j]>a[k]的对数n，m*n就是。。。

要求a[i]>a[j]的个数还是一样的，那么a[j]>a[k]的个数呢？

两种思路：

1.得到a[i]>a[j]的对数后，将数列倒过来后再求a[j]<a[k]的对数

2.更简单的做法是，找到规律发现，n = 整个数列中比a[j]小的数 — 在a[j]前面已经出现的比a[j]小的数的个数

即（假设数列是从1开始的） n = (a[j] -1) - (j - 1 - m )