Ultra-QuickSort(树状数组求逆序对数)

Ultra-QuickSort

题目链接：http://poj.org/problem?id=2299

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 51641		Accepted: 18948

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
题解：树状数组求逆序对数，就是求对于每一个数后面有多少个数字比它自己的数字小，那么从后向前遍历所有的数字，a[x]表示的是当前状态下小于等于x的值得个数.那么从后往前的扫描所有的数值统计后，再把这个数字对应的a[x]++;这样在扫描它前面的数的时候就相当于考虑这个数了，这种总是求a的前缀和和对单独点修改的操作可以使用树状数组解决。
注意这个题要解决的数很大，所以要离散化一下，这里介绍两种离散化的方法：
1.写一个二分查找的函数，先sort()一下，然后对于每个值，find(x)返回的下标值就是离散化后的结果，这里注意因为树状数组不能处理下标是0的情况，所以要将编号从1开始。而且要注意结果有可能会超int所以要用long long ,因为这个wa了好多次。
代码：

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 #define ll long long
 const ll N = ;
 ll a[N];
 ll mp[N];
 ll tree[N];
 ll lowbit(ll x){
     return x&(-x);
 }
 ll sum(ll x){
     ll ans = ;
     while(x > ){
         ans += tree[x];
         x-=lowbit(x);
     }
     return ans;
 }
 void add(ll x){
     while(x<=N){
         tree[x]++;
         x+=lowbit(x);
     }
 }
 ll n;
 ll find(ll x){
     ll l = ;
     ll r = n-;
     ll mid = (l+r)/;
     while(l<=r){
         if(a[mid]==x) return mid;
         else if(a[mid]<x) l = mid+;
         else if(a[mid]>x) r = mid-;
         mid = (l+r)/;
     }
 }
 int main()
 {
     while(~scanf("%d",&n))
     {
         if(n==) return ;
         memset(tree,,sizeof(tree));
         for(ll i = ; i < n; i++){
             scanf("%I64d",&mp[i]);
             a[i] = mp[i];
         }
         sort(a,a+n);
         ll ans = ;
         for(ll i = n-; i >= ; i--){//从后往前扫描
             mp[i] = find(mp[i])+;
             //prllf("%d \n",mp[i]);
             ans += sum(mp[i]-);
             //prllf("ans = %d ",ans);
             add(mp[i]);
         }
         printf("%I64d\n",ans);
     }
     return ;
 }