hdu 1394 Minimum Inversion Number（树状数组，线段树）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11121    Accepted Submission(s): 6842

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

 

 

求最小逆序数，因为是0到n的排列，那么如果把第一个数放到最后，对于这个数列，逆序数是减少a[i],而增加n-1-a[i]的。因为a[0]后面肯定有a[0]个比它小的数，n-1-a[0]个比它大的，每次把a[i]个当a[0],就是减少a[i],增加n-1-a[i]。

树状数组：

//time 31ms#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>using namespace std;const int maxn=10000+100;int a[maxn];int b[maxn];int low(int k){    return k&(-k);}void update(int k,int v){    while(k<maxn)    {        a[k]+=v;        k+=low(k);    }}int getsum(int k){    int ans=0;    while(k>0)    {        ans+=a[k];        k-=low(k);    }    return ans;}int main(){    int n;    while(~scanf("%d",&n))    {        int ans=0;        memset(a,0,sizeof(a));        for(int i=1;i<=n;i++)        {            scanf("%d",&b[i]);            b[i]++;//树状数组统计不到0，会陷入死循环            ans+=getsum(maxn-1)-getsum(b[i]);//前面的比它大的            update(b[i],1);        }        int Min=ans;        for(int i=1;i<=n;i++)        {            ans+=n-b[i]+1-b[i];            if(ans<Min)            Min=ans;        }        printf("%d\n",Min);    }    return 0;}

线段树：

//46ms#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int maxn=100000+1000;int a[maxn];struct node{    int l;    int r;    int sum;}edge[3*maxn];void build(int i,int l,int r){    edge[i].l=l;    edge[i].r=r;    if(l==r)    {        edge[i].sum=0;        return;    }    int mid=(l+r)>>1;    build(i<<1,l,mid);    build(i<<1|1,mid+1,r);    edge[i].sum=0;}void add(int i,int k,int v){    edge[i].sum+=v;    if(edge[i].l==edge[i].r)    {        return;    }    int mid=(edge[i].l+edge[i].r)>>1;    if(mid>=k)    {        add(i<<1,k,v);    }    else    {        add(i<<1|1,k,v);    }}int sum(int i,int l,int r){    if(edge[i].l==l&&edge[i].r==r)    return edge[i].sum;    int mid=(edge[i].l+edge[i].r)>>1;    if(r<=mid)    {        return sum(i<<1,l,r);    }    else if(l>mid)  return sum((i<<1)|1,l,r);    else return sum(i<<1,l,mid)+sum((i<<1)|1,mid+1,r);}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        int ans=0;        build(1,0,n-1);        for(int i=0;i<n;i++)          scanf("%d",&a[i]);        for(int i=0;i<n;i++)        {            ans+=sum(1,a[i],n-1);            add(1,a[i],1);        }        int Min=ans;        for(int i=0;i<n;i++)        {            ans+=n-a[i]-a[i]-1;            if(ans<Min)Min=ans;        }        printf("%d\n",Min);    }    return 0;}

    }
    return 0;
}