hdu 2689 Sort it

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2689

题目分析：求至少交换多少次可排好序，可转换为逆序对问题。 用冒泡排序较为简单，复杂度较大~~ 也可用归并排序，复杂度O(lognn), 统计个数后复杂都不变。

/*
Sort it

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2660    Accepted Submission(s): 1910


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
 

Source
ZJFC 2009-3 Programming Contest
 
*/
//冒泡排序
#include <cstdio>
const int maxn = 1000 + 10;
int a[maxn];
void swap(int i, int j)
{
    int t;
    t = a[i];
    a[i] = a[j];
    a[j] = t;
}

int main()
{
    int n;
    while(~scanf("%d", &n)){
        int cnt = 0;
        for(int i = 0; i < n; i++) scanf("%d", &a[i]);
        for(int i = 0; i < n-1; i++)
            for(int j = n-1; j >= i+1; j--){
                if(a[j] < a[j-1]){
                     swap(j, j-1);
                     cnt++;
                }
            }
        printf("%d
", cnt);
    }
    return 0;
}

//归并排序
#include <cstdio>
#include <cstring>
const int maxn = 1000 + 10;
int a[maxn], t[maxn], cnt;
void merge_sort(int x, int y)
{
    if(y-x > 1){
    int m = x + (y-x)/2;
    int p = x, q = m, i = x;
    merge_sort(x, m);
    merge_sort(m, y);
    while(p < m || q < y){
        if(q >= y || (p < m && a[p] <= a[q])) t[i++] = a[p++];
        else {
            t[i++] = a[q++];
            cnt += m-p;
        }
    }
    for(i = x; i < y; i++) a[i] = t[i];
    }
}

int main()
{
    int n;
    while(~scanf("%d", &n)){
        for(int i = 0; i < n; i++){
            scanf("%d", &a[i]);
        }
        cnt = 0;
        merge_sort(0, n);
        printf("%d
", cnt);
    }
    return 0;
}