poj 2299 Ultra-QuickSort （树状数组）

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

题意

数字很大的情况下要求逆序对。多组数据。

题解

树状数组离散化后求逆序对。（话说这题的图是不是个马桶塞啊）

代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n, a[500010], b[500010], pos[500010], tree[500010];int lowbit(int x) {    return x & (- x);}void add(int x) {    while(x <= n) {        tree[x] += 1;        x += lowbit(x);    }}long long sum(int x) {    long long ans = 0;    while(x) {        ans += tree[x];        x -= lowbit(x);    }    return ans;}int main() {    while(scanf("%d", &n) && n) {        memset(tree, 0, sizeof(tree));        for(int i = 1; i <= n; i ++)            scanf("%d", &a[i]), b[i] = a[i];        sort(b + 1, b + n + 1);        for(int i = 1; i <= n; i ++)            pos[i] = upper_bound(b + 1, b + n + 1, a[i]) - b - 1;        long long ans=0;        for(int i = 1; i <= n; i ++) {            add(pos[i]);            ans = ans + i - sum(pos[i]);        }        printf("%lld\n", ans);    }    return 0;}