HDU1394 Minimum Inversion Number

 

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

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

10
1 3 6 9 0 8 5 7 4 2

Sample Output

16

Source

ZOJ Monthly, January 2003

 

在环上找一个起点，使得从该起点数起的长度为n的链中逆序对数最小。

 

破环为链会T。

正解是每次将链头的数移到尾部，用公式计算出逆序对的变化量。

 

WA了三次才看到题目包含多组数据，WA的一声就哭了

 

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int mxn=12000;
int t[mxn];
int n;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x;
}
int lowbit(int x){
    return x&-x;
}
void add(int pos,int v){
    while(pos<=n){
        t[pos]+=v;
        pos+=lowbit(pos);
    }
    return;
}
int ask(int x){
    int res=0;
    while(x){
        res+=t[x];
        x-=lowbit(x);
    }
    return res;
}
int ans=1e8;
int a[mxn];
int main(){
    while(scanf("%d",&n)!=EOF){
        memset(t,0,sizeof t);
        ans=1e8;
        int i,j;
        int res=0;
        for(i=1;i<=n;i++){
            a[i]=read();
            a[i]++;
            res+=ask(n)-ask(a[i]);
            add(a[i],1);
        }
        ans=min(ans,res);
        for(i=1;i<=n;i++){
            res+=(n-a[i])-(a[i]-1);
            ans=min(ans,res);
        }
        printf("%d
",ans);
    }
    return 0;
}