poj2299 Ultra-QuickSort
Ultra-QuickSort
| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 50361 | Accepted: 18458 |
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
离散化+树状数组求逆序对
(离散化用map会超时…)
#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<map>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define LL long long
#define MAXN 500005
#define pa pair<int,int>
using namespace std;
int n,b[MAXN],f[MAXN];
LL ans;
struct data
{
int val,pos;
}a[MAXN];
int read()
{
int ret=0,flag=1;char ch=getchar();
while (ch<'0'||ch>'9'){if (ch=='-') flag=-1;ch=getchar();}
while (ch>='0'&&ch<='9'){ret=ret*10+ch-'0';ch=getchar();}
return ret*flag;
}
void add(int k)
{
for(int i=k;i<=n;i+=i&(-i)) f[i]++;
}
LL getsum(int k)
{
LL ret=0;
for(int i=k;i>0;i-=i&(-i)) ret+=f[i];
return ret;
}
bool cmp(data x,data y)
{
return x.val<y.val;
}
int main()
{
n=read();
while (n)
{
memset(f,0,sizeof(f));
ans=0;
F(i,1,n) a[i].val=read(),a[i].pos=i;
sort(a+1,a+n+1,cmp);
F(i,1,n)
{
if (i>1&&a[i].val==a[i-1].val) b[a[i].pos]=b[a[i-1].pos];
else b[a[i].pos]=i;
}
D(i,n,1)
{
if (b[i]>1) ans+=getsum(b[i]-1);
add(b[i]);
}
printf("%lld\n",ans);
n=read();
}
}