【POJ2266】【树状数组+离散化】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

【分析】

开始离散化用MAP T了半天，不活了..

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <algorithm>

 4 #include <cstring>

 5 #include <vector>

 6 #include <utility>

 7 #include <iomanip>

 8 #include <string>

 9 #include <cmath>

10 #include <map>

11 

12 const int MAXN = 500000 + 10;

13 const int MAXM = 500000 + 10;

14 //const int MAXM = 2000 + 10;

15 const int MAXL = 10;

16 using namespace std;

17 struct DATA{

18        int val;

19        int order;

20        bool operator < (DATA b)const{

21             return val < b.val;

22        }

23 }rem[MAXN];

24 typedef long long ll;

25 int n;

26 int data[MAXN];

27 int C[MAXN];

28 

29 void init(){

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

31          scanf("%d", &data[i]);

32          C[i] = 0;

33          rem[i].val = data[i];

34          rem[i].order = i;

35      }

36      sort(rem + 1, rem + 1 + n);

37      for (int i = 1; i <= n; i++) data[rem[i].order] = i;

38      //for (int i = 1; i <= n; i++) printf("%d\n", data[i]);printf("\n");

39 }

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

41 int sum(int x){

42    int cnt = 0;

43    while (x > 0){

44          cnt += C[x];

45          x -= lowbit(x);

46    }

47    return cnt;

48 }

49 void add(int x){

50    while (x <= n){

51          C[x]++;

52          x += lowbit(x);

53    }

54    return;

55 }

56 

57 void work(){

58      ll Ans = 0;

59      //前面共 i - 1个数字 

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

61          Ans += (i - 1 - sum(data[i]));//严格大于

62          add(data[i]); 

63      }

64      printf("%lld\n", Ans);

65 }

66 

67 int main(){

68      #ifdef LOCAL

69      freopen("data.txt", "r", stdin);

70      freopen("out.txt", "w", stdout); 

71      #endif 

72      while (scanf("%d", &n) && n){

73            init();

74            work();

75      }

76      return 0;

77 }

78  

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