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HDU 1394Minimum Inversion Number 数状数组 逆序对数量和

2024-08-15 17:13:19   220人阅读

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18543    Accepted Submission(s): 11246


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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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1394
题意：求出对(ai,aj)(i<j,ai>aj)的全部数量。即求ai右边比ai小的数的个数和。每次变换a的序列，求出最小的逆序对数量和。
思路： 因为树状数组的最基本功能就是求比某点 x 小的点的个数。所以逆向存储ai。
代码：
技术分享
#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int MAXN=1e6+100,INF=1e9+100;int a[MAXN],c[MAXN],ans[MAXN];int lowbit(int x){    return x&(-x);}void add(int i,int val){    for(i; i<=MAXN; i+=lowbit(i))        c[i]+=val;}int sum(int i){    int s=0;    for(i; i>0; i-=lowbit(i))        s+=c[i];    return s;}int main(){    int i,j,t,n;    while(scanf("%d",&n)!=EOF)    {        for(i=n; i>=1; i--)            scanf("%d",&a[i]);        memset(c,0,sizeof(c));        memset(ans,0,sizeof(ans));        int cou=0;        for(i=1; i<=n; i++)        {            ans[i]=sum(a[i]+1);            cou+=ans[i];            add(a[i]+1,1);        }        int Min=cou;        for(i=n; i>1; i--)        {            for(j=1; j<=n; j++)            {                if(j==i) continue;                if(a[i]<a[j])                {                    cou++;                    ans[j]++;                }            }            cou-=ans[i];            ans[i]=0;            if(cou<Min) Min=cou;        }        cout<<Min<<endl;    }    return 0;}
逆序对

 

HDU 1394Minimum Inversion Number 数状数组 逆序对数量和


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