HDU 1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17988    Accepted Submission(s): 10927

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

 

树状数组更新单点值，求区间和。这里用到了一个数学知识：设ans为逆序和，a[i]输入的序列，则将a[i]移至队尾后，序列的逆序数变为ans=ans-a[i]+n-a[i]-1；在求出初次输入的逆序数后，进行一个 1~n for循环即可求解。代码如下：

Problem : 1394 ( Minimum Inversion Number )     Judge Status : Accepted
RunId : 18054217    Language : C++    Author : 15030120010
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta

#include<stdio.h>#include<string.h>int tree[5005],a[5005];int n;int lowbit(int x){ return x&-x;}int query(int x){ int s=0; while(x>0) { s+=tree[x]; x-=lowbit(x); } return s;}void add(int x,int v){ while(x<=n) { tree[x]+=v; x+=lowbit(x); }}int main(){ int i,x,ans; while(scanf("%d",&n)!=EOF) { memset(tree,0,sizeof(tree)); memset(a,0,sizeof(a)); ans=0; for(i=1;i<=n;++i) { scanf("%d",&a[i]); ans+=query(n)-query(a[i]+1),add(a[i]+1,1); } x=ans; for(i=1;i<=n;++i) { ans+=n-a[i]-a[i]-1; if(x>ans) x=ans; } printf("%d\n",x); } return 0;}

#include<stdio.h>#include<string.h>int tree[5005],a[5005];int n;int lowbit(int x){    return x&-x;}int query(int x){    int s=0;    while(x>0)    {        s+=tree[x];        x-=lowbit(x);    }    return s;}void add(int x,int v){    while(x<=n)    {        tree[x]+=v;        x+=lowbit(x);    }}int main(){    int i,x,ans;    while(scanf("%d",&n)!=EOF)    {        memset(tree,0,sizeof(tree));        memset(a,0,sizeof(a));        ans=0;        for(i=1;i<=n;++i)        {            scanf("%d",&a[i]);            ans+=query(n)-query(a[i]+1),add(a[i]+1,1);        }        x=ans;        for(i=1;i<=n;++i)        {            ans+=n-a[i]-a[i]-1;            if(x>ans) x=ans;        }        printf("%d\n",x);    }    return 0;}