1394 ( Minimum Inversion Number ) (线段树)

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): 16259    Accepted Submission(s): 9891

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

 

线段树水题，输入一个x，查询【x，n-1】区间有多少数，然后update【x】，如果先更新，那么查询的值就是【x+1，n-1】了，可能会出错

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <map>#include <set>#include <stack>#include <string>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long longusing namespace std;const int SIZE=5e3+10;const int maxn=500000;const int Mod=1e8;int sum[SIZE<<2];int a[SIZE];void pushup(int rt){    sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void build(int l,int r,int rt){    memset(sum,0,sizeof(sum));}void update(int x,int l,int r,int rt){    if(l==r){        sum[rt]=1;        return;    }    int m=(l+r)>>1;    if(x<=m)update(x,lson);    else update(x,rson);    pushup(rt);}int query(int x,int l,int r,int rt){    if(x<=l){        return sum[rt];    }    int m=(l+r)>>1;    int cnt=0;    if(x<m)cnt+=query(x,lson)+query(m+1,rson);    else cnt+=query(x,rson);    return cnt;}int main(){    int n;    while(scanf("%d",&n)!=EOF){        int sum=0;        build(0,n-1,1);        for(int i=0;i<n;i++){            scanf("%d",&a[i]);            sum+=query(a[i],0,n-1,1);            update(a[i],0,n-1,1);        }        int ans=sum;        for(int i=0;i<n;i++){            sum+=n-1-2*a[i];            ans=min(ans,sum);        }        printf("%d\n",ans);    }    return 0;}