POJ 2299 Ultra-QuickSort（树状数组）

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

题意：求按冒泡升序排序的交换次数。

题解：按顺序依次插入树状数组里，每次统计a[i]前面有多少个数，然后ans+=i-num;

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#define N 5000010
#define ll long long

using namespace std;

int n;
struct node {
    int x;
    int num;
} a[N];
int bit[N];

bool cmp_1(node a,node b) {
    return a.x<b.x;
}

bool cmp_2(node a,node b) {
    return a.num<b.num;
}

int sum(int i) {
    int s=0;
    while(i>0) {
        s+=bit[i];
        i-=i&-i;
    }
    return s;
}

void add(int i,int x) {
    while(i<=n) {
        bit[i]+=x;
        i+=i&-i;
    }
}
int main() {
    //freopen("test.in","r",stdin);
    while(cin>>n&&n) {
        for(int i=1; i<=n; i++) {
            scanf("%d",&a[i].x);
            a[i].num=i;
        }
        sort(a+1,a+n+1,cmp_1);
        for(int i=1; i<=n; i++)//离散化
            a[i].x=i;
        sort(a+1,a+n+1,cmp_2);
        memset(bit,0,sizeof bit);
        ll ans=0;
        for(int i=1; i<=n; i++) {
            ans+=i-sum(a[i].x)-1;
            add(a[i].x,1);
        }
        printf("%lld\n",ans);
    }
    return 0;
}

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