HDU 1394 Minimum Inversion Number

HDU 1394 Minimum Inversion Number

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

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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 

解题思路：
1，题目的意思是给你一个n，然后是一段包含0~n-1全部数字的序列，然后有一个操作
把前面的数放到后面来，这样的操作可以形成n个序列，让你求出这样所有的逆序序列里最小的逆序数
2，一个数列逆序数的求法有归并排序（不介绍）然后就是树状数组了
3，树状数组实现求取逆序数：把树状数组当做标记数组使用，从后往前输入每一个数据计算出现了多少个
比这个数小的数的个数因为树状数组可以快速求和，这样就可以在nlogn的时间复杂度内求得逆序数了
4，然后就是怎么根据第一个求得的下一个状态的逆序数了
5，对于开头的第一个数，如果这个数放在后面，首先逆序数会减少【后面有多少个比他小的数的个数】，
加到后面会导致逆序数增加【前面有多少个比他大的】，由于这里包含了所有的0~n-1的数，所以很容易求得所有的逆序数。
6，然后简单的代码实现就可以了。


       
///树状数组求逆序数
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 5005 ;
int C[4*maxn] ;
int n ;///节点总数
int lowbit(int x){return x&-x;}
int sum(int x){
    int ret = 0 ;
    while(x>0){
        ret+=C[x];
        x-=lowbit(x) ;
    }
    return ret ;
}
void add(int x,int d){
    while(x<=n){
        C[x]+=d ;
        x+=lowbit(x) ;
    }
    return ;
}
int num[maxn] ;
int x[maxn] ;
int main(){
    while(~scanf("%d",&n)){
        memset(C,0,sizeof(C)) ;
        memset(num,0,sizeof(num)) ;
        for(int i=0;i<n;i++){
            scanf("%d",&x[i]);
            x[i]++ ;
        }
        //printf("yes\n");
        int ans = 0 ;
        for(int i=n-1;i>=0;i--){
            //printf("i = %d\n",i);
            num[i] = sum(x[i]);
            ans+=num[i] ;
            //printf("ans = %d\n",ans);
            add(x[i],1) ;
        }
        //printf("ans=%d\n",ans);
        //printf("yes2\n");
        int temp = ans ;
        for(int i=0;i<n;i++){
            temp=temp-(x[i]-1)+(n-x[i]);
            ans = min(temp,ans);
        }
        printf("%d\n",ans);
    }
    return 0;
}