POJ 2299（树状数组，离散化）

Ultra-QuickSort
Time Limit: 7000MS Memory Limit: 65536K
Total Submissions: 52711 Accepted: 19317
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
Source


Waterloo local 2005.02.05

题意：求逆序数

题解：可以将数字插入树状数组，每次统计比单前数字小的个数，我们将每个数字的权值设置为1，那么只要求在他前面数字的和就知道了，对应的逆序数格个数是i-sum(a[i]),这里使用树状 数组维护区间和

这一题由于数字范围过大，内存无法承受如此大的空间，我们可以看到n最多才5e5，所以最多有n个不同数字，离散化即可，这一题很诡异的是卡STL。。。。。可以使用数组去映射，记录下每个数字的原来位置，将数字按升序排序，这样再将原来的位置依次插入i，这样相对的大小规则不会发生变化，求得的逆序数也不会发生变化，这样就达到了节省空间的目的

#include<cstdio>  #include<cstring>  #include<cstdlib>  #include<cmath>  #include<iostream>  #include<algorithm>  #include<vector>  #include<map>  #include<set>  #include<queue>  #include<string>  #include<bitset>  #include<utility>  #include<functional>  #include<iomanip>  #include<sstream>  #include<ctime>  using namespace std;#define N int(6e5+10)  #define inf int(0x3f3f3f3f)  #define mod int(1e9+7)  typedef long long LL;int c[N], n;int a[N];struct  point{int x, y;bool operator<(const point &x1)const{return this->x<x1.x;}}node[N];inline void add(int x){for (int i = x; i <= n; i += (i&(-i))){c[i]++;}}inline int sum(int x){int res = 0;for (int i = x; i>0; i -= (i&(-i))){res += c[i];}return res;}int main(){#ifdef CDZSC  freopen("i.txt", "r", stdin);//freopen("o.txt","w",stdout);  int _time_jc = clock();#endif  int  x, y;while (~scanf("%d", &n)){if (n <= 0)break;LL ans = 0;for (int i = 1; i<=n; i++){c[i] = 0;scanf("%d", &node[i].x);node[i].y = i;}sort(node + 1, node + 1 + n);for (int i = 1; i <= n; i++){a[node[i].y] = i;c[i] = 0;}for (int i = 1; i <= n; i++){add(a[i]);ans += (i - sum(a[i]));}printf("%lld\n", ans);}return 0;}