UVA 10810 - Ultra-QuickSort（树状数组+离散化求逆序对)

Problem B: Ultra-QuickSort

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of ndistinct 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.

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.

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

59105431230

Output for Sample Input

60

题意：问排序好需要交换几次。

思路：利用树状数组求逆序对。。但是数字很大的时候就需要离散化。

代码：

#include <stdio.h>#include <string.h>#include <map>#include <algorithm>using namespace std;const int N = 500005;int n, num[N], save[N], bit[N];map<int, int> hash;void add(int x, int v) {while (x <= n) {bit[x] += v;x += (x&(-x));}}int get(int x) {int ans = 0;while (x > 0) {ans += bit[x];x -= (x&(-x));}return ans;}void init() {hash.clear();memset(bit, 0, sizeof(bit));for (int i = 1; i <= n; i++) {scanf("%d", &num[i]);save[i] = num[i];}sort(save + 1, save + n + 1);for (int j = 1; j <= n; j++)hash[save[j]] = j;}long long solve() {long long ans = 0;for (int i = 1; i <= n; i++) {ans += get(n) - get(hash[num[i]]);add(hash[num[i]], 1);}return ans;}int main() {while (~scanf("%d", &n) && n) {init();printf("%lld\n", solve());}return 0;}