【数列上的分治】——hdu2689

                                       Sort it

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3938    Accepted Submission(s): 2791

Problem Description

You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need.
For example, 1 2 3 5 4, we only need one operation : swap 5 and 4.

 

 

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 <= 1000); the next line contains a permutation of the n integers from 1 to n.

 

 

Output

For each case, output the minimum times need to sort it in ascending order on a single line.

 

 

Sample Input

3

1 2 3

4

4 3 2 1

 

 

Sample Output

0

6

 

 

Author

WhereIsHeroFrom

 

题意：求逆序数

 

思路：把数列分成两半

        则逆序对变成了三种：

       （1）i，j都在b内

       （2）i，j都在c内

       （3）i，j一个在b内，一个在c内

前两种：递归的时候就包含在上一层的cnt内了

第三种：归并排序的时候计算即可。

 

附上代码：

//利用归并排序
#include<iostream>
#include<vector>
using namespace std;

typedef long long ll;

ll merge_count(vector<int> &a)
{
    if(a.size()<=1)return 0;

    ll cnt=0;

    vector<int> b(a.begin()            , a.begin()+a.size()/2);
    vector<int> c(a.begin()+a.size()/2 , a.end()             );

    cnt+=merge_count(b);//（1）
    cnt+=merge_count(c);//（2）
    //此时，b，c已经排好序了

    //归并排序          //（3）
    int ai=0,bi=0,ci=0;
    while(ai<a.size())
    {
        //非逆序
        if(bi<b.size() && (ci==c.size() || b[bi]<=c[ci]) )
        {
            a[ai++]=b[bi++];
        }
        //逆序
        else
        {
            cnt+=a.size()/2 - bi;
            a[ai++]=c[ci++];
        }
    }
    return cnt;
}

int main()
{
    int t;
    while(cin>>t)
    {
        vector<int>A;//注意vector初始化的位置
        int num;
        for(int i=0;i<t;i++)
        {
            cin>>num;
            A.push_back(num);
        }
        cout<<merge_count(A)<<endl;
    }
    return 0;
}