Ultra-QuickSort 树状数组+离散化；

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 66796		Accepted: 25021

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. InputThe 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.OutputFor 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 Input5
9
1
0
5
4
3
1
2
3
0
Sample Output6
0
             这里要注意的两个点就是离散化操作和怎样理解   ans += s - getsum(c[s]) + 1;这行代码所表达的内容；
             还有就是auto是c++ 14的内容，现在国内大多数oj是不支持的；（https://vjudge.net/problem/EOlymp-1303 在这里可以用auto）
             现在网上许多关于这道题的树状数组解答，基本没考虑过n个数中有同样的情况，他们大多是按照互不相同的n个数的条件来做的，可是这样考虑是不全面的，下面就是关于有相同的数的情况下的离散化操作；
             树状数组c开始的时候都初始化为0的，所以按照 9 1 0 5 4输入时，在进行离散化操作后也就会按照5 2 1 4 3的形式进行计算，5输入时，其他全是0，这时候getsum（5）=1；也就是说这时候不比5大的数的的数量之和是1（自己），因他而产生的逆序数就是0了；到输入4（离散化的4）的时候，c[1]=1,c[2]=2,c[3]=0,c[4]=1,所以getsum（4）=3，也就是在他之前的位置的不大于他的数的数量总和为3；这样依次进行计算即可。#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring>
#include<cstdio>
using namespace std;
long long int m[500001];
int c[500001], n;
int tree[500001];
int lowbit(int x)
{
return x&(-x);
}
int getsum( int x)
{
long long int ans = 0;
for (long long int s = x; s >= 1; s -= lowbit(s))
{
// cout << s << endl;
ans += tree[s];
// cout << s << " " << tree[s] << endl;
// cout << ans << endl;
}
return ans;
}
void add( int x, int d)
{
for (long long int s = x; s <= n; s+=lowbit(s))
{
tree[s] += d;
}
}
vector < long long int > q;
void init()
{
memset(tree, 0, sizeof(tree));
memset(c, 0, sizeof(c));
memset(m, 0, sizeof(m));
q.clear();
}
int main()
{
while (cin >> n&&n)
{
init();
for (int s = 0; s < n; s++)
{
cin >> m[s];
q.push_back(m[s]);
}
sort(q.begin(), q.end());
auto siz = unique(q.begin(), q.end());
for (int s = 0; s < n; s++)
{
c[s] = lower_bound(q.begin(), siz, m[s]) - q.begin() + 1;
}
long long int ans = 0;
for (int s = 0; s < n; s++)
{
add(c[s], 1);
ans += s - getsum(c[s]) + 1;
// cout << s + 1 - getsum(c[s]) << endl;
}
printf("%lld\n", ans);
}
return 0;
}