xtu数据结构 C. Ultra-QuickSort

C. Ultra-QuickSort

Time Limit: 7000ms

Memory Limit: 65536KB

64-bit integer IO format:  %lld      Java class name: Main

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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

解题：求逆序数，归并排序或者快排+树状数组都可以。坑爹的地方在于要使用long long ,害我WA了几次。逗比。。。。。。

树状数组+快速排序

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #define LL long long

13 #define INF 0x3f3f3f

14 using namespace std;

15 const int maxn = 500002;

16 struct node {

17     int val,index;

18 } p[maxn];

19 LL tree[maxn];

20 bool cmp(const node &x,const node &y) {

21     return x.val > y.val;

22 }

23 int lowbit(int x) {

24     return x&(-x);

25 }

26 void update(int x,int val) {

27     for(int i = x; i < maxn; i += lowbit(i)) {

28         tree[i] += val;

29     }

30 }

31 LL sum(int x) {

32     LL ans = 0;

33     for(int i = x; i; i -= lowbit(i)) {

34         ans += tree[i];

35     }

36     return ans;

37 }

38 int main() {

39     int n,i;

40     LL ans;

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

42         for(i = 0; i < n; i++) {

43             scanf("%d",&p[i].val);

44             p[i].index = i+1;

45         }

46         sort(p,p+n,cmp);

47         memset(tree,0,sizeof(tree));

48         int pre = INT_MIN;

49         for(ans = i = 0; i < n; i++) {

50             update(p[i].index,1);

51             ans += sum(p[i].index-1);

52 

53         }

54         printf("%lld\n",ans);

55     }

56     return 0;

57 }

View Code

归并排序

 1 #include <cstdio>

 2 #define LL long long

 3 LL sum,dt[500010];

 4 void mysort(int lft,int rht,int step){

 5     static LL temp[500010];

 6     int md = lft + (step>>1),i = 0,k = 0,j = 0;

 7     while(lft + i < md && md + j < rht){

 8         if(dt[lft+i] > dt[md+j]){temp[k++] = dt[md+j];j++;

 9         }else{temp[k++] = dt[lft+i];i++;sum += j;}

10     }

11     while(lft+i < md){temp[k++] = dt[lft+i];i++;sum += j;}

12     while(md+j < rht){temp[k++] = dt[md+j];j++;}

13     for(i = 0; i < k; i++) dt[lft+i] = temp[i];

14 }

15 void ms(int n){

16     int len = 1,step = 2,m,i,u,v;

17     sum = 0;

18     while(len < n){len <<= 1;}

19     m = len/step;

20     while(step <= len){

21         for(i = 0; i < m; i++){

22             u = i*step;v = (i+1)*step;

23             mysort(u,v>n?n:v,step);

24         }

25         step <<= 1;m = len/step;

26     }

27 }

28 int main(){

29     int n,i;

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

31         for(i = 0; i < n; i++)

32             scanf("%d",dt+i);

33         ms(n);

34         printf("%lld\n",sum);

35     }

36     return 0;

37 }

View Code