poj 2299

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 66949		Accepted: 25081

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

因为数组里元素比较大,所以需要离散化处理

 

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <string.h>
#include <cstring>
#include <stdio.h>
#define lowbit(x) x&(-x)

using namespace std;
const int maxn=5e5+10;
typedef long long ll;

int n;
int a[maxn];
ll c[maxn];

struct Node{
    int v;
    int order;
}b[maxn];

void add(int x){
    for(int i=x;i<=n;i+=lowbit(i))
        c[i]+=1;
}

int getsum(int x){
    ll as=0;
    for(int i=x;i>0;i-=lowbit(i)){

        as+=c[i];
    }
    return as;
}
bool cmp(const Node&a,const Node&b){
    return a.v<b.v;
}

int main(){
    ios::sync_with_stdio(0);
    int i,j;
    while(~scanf("%d",&n)&&n){
        for(i=1;i<=n;i++){
            scanf("%d",&b[i].v);
            b[i].order=i;
        }
        sort(b+1,b+n+1,cmp);
        for(i=1;i<=n;i++){
            a[b[i].order]=i;
        }
        ll  ans=0;
        memset(c,0,sizeof(c));
        for(i=1;i<=n;i++){
            add(a[i]);
            ans+=(i-getsum(a[i]));
        }
        cout<<ans<<endl;
    }
    return 0;
}