poj 2299 Ultra-QuickSort

题目连接

http://poj.org/problem?id=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.

SampleInput

5
9
1
0
5
4
3
1
2
3
0

SampleOutput

6

0

用线段树求逆序数。。

 1 #include<algorithm>

 2 #include<iostream>

 3 #include<cstdlib>

 4 #include<cstring>

 5 #include<cstdio>

 6 #include<vector>

 7 #include<map>

 8 #include<set>

 9 using std::cin;

10 using std::cout;

11 using std::endl;

12 using std::find;

13 using std::sort;

14 using std::set;

15 using std::map;

16 using std::pair;

17 using std::vector;

18 using std::multiset;

19 using std::multimap;

20 #define all(c) (c).begin(), (c).end()

21 #define iter(c) decltype((c).begin())

22 #define cls(arr,val) memset(arr,val,sizeof(arr))

23 #define cpresent(c, e) (find(all(c), (e)) != (c).end())

24 #define rep(i, n) for (int i = 0; i < (int)(n); i++)

25 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)

26 #define pb(e) push_back(e)

27 #define mp(a, b) make_pair(a, b)

28 #define lc (root<<1)

29 #define rc (root<<1|1)

30 #define mid ((l+r)>>1)

31 const int Max_N = 500100;

32 typedef unsigned long long ull;

33 struct Node { int val; };

34 struct P {

35     int v, id;

36     friend bool operator<(const P &a, const P &b) {

37         return a.v == b.v ? a.id < b.id : a.v < b.v;

38     }

39 }rec[Max_N];

40 struct SegTree {

41     Node seg[Max_N << 2];

42     inline void init() {

43         cls(seg, 0);

44     }

45     inline void insert(int root, int l, int r, int p) {

46         if (p > r || p < l) return;

47         if (p <= l && p >= r) { seg[root].val++; return; }

48         insert(lc, l, mid, p);

49         insert(rc, mid + 1, r, p);

50         seg[root].val = seg[lc].val + seg[rc].val;

51     }

52     inline ull query(int root, int l, int r, int x, int y) {

53         if (x > r || y < l) return 0;

54         if (x <= l && y >= r) return seg[root].val;

55         ull ret = 0;

56         ret += query(lc, l, mid, x, y);

57         ret += query(rc, mid + 1, r, x, y);

58         return ret;

59     }

60 }seg;

61 int main() {

62 #ifdef LOCAL

63     freopen("in.txt", "r", stdin);

64     freopen("out.txt", "w+", stdout);

65 #endif

66     int n;

67     while (~scanf("%d", &n) && n) {

68         seg.init();

69         ull res = 0;

70         for (int i = 1; i <= n; i++) scanf("%d", &rec[i].v), rec[i].id = i;

71         sort(rec + 1, rec + n + 1);

72         for (int i = 1; i <= n; i++) {

73             res += seg.query(1, 1, n, rec[i].id, n);

74             seg.insert(1, 1, n, rec[i].id);

75         }

76         printf("%lld\n", res);

77     }

78     return 0;

79 }

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