应用归并排序（求逆序数）或树状数组

Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence 
9 1 0 5 4 ,
Ultra-QuickSort produces the output 
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0
大意：求冒泡排序交换的个数，也就是求逆序数，套一下模板

#include<cstring>
#include<cstdio>
#include<algorithm>
const int MAX = 510000;
int a[MAX],temp[MAX];
int ans = 0;
void mergesort(int *a,int first,int mid,int last,int *temp)
{
   int i = first,j = mid + 1;
   int  k = first;
    while( i <= mid && j <= last){
          if(a[i] <= a[j])
          temp[k++] = a[i++];
          else {
            ans += j - k;
           temp[k++] = a[j++];
          }
    }
    while( i <= mid)
        temp[k++] = a[i++];
    while( j <= last)
        temp[k++] = a[j++];
    for(int i = first;i <= last;i++)
        a[i] = temp[i];
}
void mergearray(int *a,int first,int last,int *temp)
{
     if(first < last){
            int mid = (first + last) >> 1;
            mergearray(a,first,mid,temp);
            mergearray(a,mid+1,last,temp);
            mergesort(a,first,mid,last,temp);
     }
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
    scanf("%d",&a[i]);
    mergearray(a,1,n,temp);
    printf("%d",ans);
  return 0;
}

 树状数组：

/************************************************
* Author        :Powatr
* Created Time  :2015-8-13 19:02:01
* File Name     :poj2299Ultra-QuickSort.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 5e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct edge{
    long long  m;
    int id;
    bool operator <(const edge &a) const{
        return m < a.m;
    }
}a[MAXN];

int C[MAXN];
int b[MAXN];
int query(int x)
{
    int ret = 0;
    while(x > 0){
        ret += C[x];
        x -= x&-x;
    }
    return ret;
}

void update(int x) { 
    while(x < MAXN){
        C[x]++;
        x += x&-x;
    }
}
int main(){
    int n;
    int m;
    while(~scanf("%d", &n)&&n){
        int cnt = 0;
        for(int i = 1; i <= n; i++){
            scanf("%I64d", &a[i].m);
            a[i].id = i;
        }
        sort(a+1, a+n+1);
        for(int i = 1; i <= n; i++)
            b[a[i].id] = i;
        ll ans = 0;
        memset(C, 0, sizeof(C));
        for(int i = 1; i <= n; i++){
            ans += i - query(b[i]) - 1;
            update(b[i]);
        }
        printf("%I64d\n", ans);
    }
    return 0;
}