poj 2299 Ultra-QuickSort

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

 

题意为给你一串数组，问用冒泡排序多少次可让数组有序（从小到大）

分析题目可知，要想知道次数需算出每个点前有多少个比他大，不妨用树状数组做

思路：先排序用离散化储存，在按顺序把每个点加入树状数组中，然后查询它前面有多少比他小的数，用i减去就是比他大的数。。（我承认讲不清。。。

挂个连接自己看 https://blog.csdn.net/jk_chen_acmer/article/details/79347003

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;

const int maxn = 5e5+10;

int n,tree[maxn],id[maxn];

struct node{
    int val,pos;
}in[maxn];

bool cmp(node a,node b){
    return a.val<b.val;
}

int lowbit(int x){
    return x&(-x);
}

void updata(int x,int val){
    while(x<=n){
        tree[x]+=val;
        x+=lowbit(x);
    }
}

int getsum(int x){
    int tot=0;
    while(x>0){
        tot+=tree[x];
        x-=lowbit(x);
    }
    return tot;
}

int main(){
       // freopen("in.txt", "r", stdin);
    while(scanf("%d",&n)!=EOF&&n){
        for(int i=1;i<=n;i++){
            scanf("%d",&in[i].val);in[i].pos=i;
        }
        sort(in+1,in+1+n,cmp);
        for(int i=1;i<=n;i++){
            id[in[i].pos]=i;
        }
        memset(tree,0,sizeof(tree));
        long long ans=0;
        for(int i=1;i<=n;i++){
            updata(id[i],1);
            ans+=(i-getsum(id[i]));
        }
        printf("%lld
",ans);
    }
    return 0;
}