Inversion Number

Description

There is a permutation P with n integers from 1 to n. You have to calculate its inversion number, which is the number of pairs of Pi and Pj satisfying the following conditions: iPj. 

Input

 The input may contain several test cases.

In each test case, the first line contains a positive integer n (n<=100000) indicating the number of elements in P. The following n lines contain n integers making the permutation P.

Input is terminated by EOF.

Output

 For each test case, output a line containing the inversion number.

题目解释：对于一个无序数组，求解逆序对的个数

解题思路：虽然说使用2个for循环就能够解决这个问题，但是使用for循环会导致超时。因此采用分治算法，将求解一个大数组的逆序数变成求解若干个小数组的逆序数。这个算法跟之前的最近邻点对的求解很类似。解题步骤如下：

（1）使用递归方式不断划分数组，直到不能再分

（2）对于相邻两个数组求解逆序数

（3）对求解完逆序数的两数组进行排序合并

（4）对于合并后的数组进行同样的操作。

后记：本来是想使用new和delete进行动态内存分配的，但是不能通过Sicily，不知道为何~

#include 
#include 
using namespace std;
int P[100005];
int element_num;
long long result;    // 输出结果的类型一定要是long long型，否则会出错。
int getIndex(int *arr, int begin, int end, int target){
    int mid = 0;
    while (begin <= end) {
        mid = (begin + end) /2;
        if (target < arr[mid]) end = mid - 1;
        else if(target > arr[mid]) begin = mid + 1;
        else return mid;
    }
    return begin;
}
void sortmerge(int *arr, int begin, int mid, int end){
    int i, j;
    int l1 = mid - begin + 1;
    int l2 = end - mid;
    int *L = (int*)malloc(sizeof(int)*l1);   // 动态创建数组存储左边数据
    int *R = (int*)malloc(sizeof(int)*l2);   // 动态创建数组存储右边数据
    for (i = 0; i < l1; i++) {
        L[i] = arr[i+begin];
    }
    for (i = 0; i < l2; i++) {
        R[i] = arr[mid+1+i];
    }
    for(i = 0; i < l2; i++){
        result += (l1 - getIndex(L, 0, l1-1, R[i]));
    }
    i = j = 0;
    int k = begin;
    while (i < l1 && j < l2) {   // 对于两部分的数据进行排序
        if (L[i] < R[j]) {
            arr[k] = L[i];
            i++;
        }
        else{
            arr[k] = R[j];
            j++;
        }
        k++;
    }
    while (i < l1) {
        arr[k] = L[i];
        k++;
        i++;
    }
    while (j < l2) {
        arr[k] = R[j];
        k++;
        j++;
    }
    free(L);
    free(R);
}
void Merge(int *arr, int begin, int end){
    if (begin < end) {
        int mid = (begin + end) / 2;
        Merge(arr, begin, mid);
        Merge(arr, mid + 1, end);
        sortmerge(arr, begin, mid, end);
    }
}
int main(){
    while (cin >> element_num) {
        for (int i = 0; i < element_num; i++) {
            cin >> P[i];
        }
        result = 0;
        Merge(P, 0, element_num-1);
        cout << result << endl;
    }
    return 0;
}