[HDU] 1394 Minimum Inversion Number [线段树求逆序数]

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11788    Accepted Submission(s): 7235

Problem Description

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

 

 

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

 

 

Output

For each case, output the minimum inversion number on a single line.

 

 

Sample Input

10

1 3 6 9 0 8 5 7 4 2

 

 

Sample Output

16

 

 

Author

CHEN, Gaoli

 

 

Source

ZOJ Monthly, January 2003

 

 

Recommend

Ignatius.L

 

题解：这题的维护信息为每个数是否已经出现。每次输入后，都从该点的值到n-1进行查询，每次发现出现了一个数，由于是从该数的后面开始找的，这个数肯定是比该数大的。那就是一对逆序数，然后逆序数+1.最后求完所有的逆序数之后，剩下的就可以递推出来了。因为假如目前的第一个数是x，那当把他放到最后面的时候，少的逆序数是本来后面比他小的数的个数。多的逆序数就是放到后面后前面比他大的数的个数。因为所有数都是从0到n-1.所以比他小的数就是x，比他大的数就是n-1-x。这样的话每次的逆序数都可以用O(1)的时间计算出来。然后找最小的时候就可以了。

 

 1 #include<cstdio>

 2 #include<algorithm>

 3 

 4 using namespace std;

 5 

 6 #define  lson  l , m , rt << 1

 7 #define  rson  m + 1 , r , rt << 1 | 1

 8 

 9 const int maxn = 5000 + 311;

10 int sum[maxn<<2];

11 

12 

13 void PushUp(int rt) 

14 {

15     sum[rt]=sum[rt<<1]+sum[rt<<1|1];

16 }

17 

18 void build(int l,int r,int rt)

19 {

20     int m;

21     

22     sum[rt]=0;

23     if(l==r) {

24         return;

25     }

26     m=(l+r)>>1;

27     build(lson);

28     build(rson);

29 }

30 

31 int query(int L,int R, int l, int r,int rt)

32 {

33     int m,ret;

34     

35     if(L<=l && r<=R) {

36         return sum[rt];

37     }

38     

39     m=(l+r) >> 1;

40     ret=0; 

41     if(L<=m) ret+=query(L,R,lson);

42     if(R>m) ret+=query(L,R,rson);

43     

44     return ret;

45 } 

46 

47 void Updata(int p,int l,int r,int rt)

48 {

49     int m;

50     if (l==r) {

51         sum[rt]++;

52         return ;

53     }

54     m=(l+r)>>1;

55     if(p<=m) Updata(p,lson);

56     else Updata(p,rson);

57     

58     PushUp(rt); 

59 }

60 

61 int main()

62 {

63     int n,sum,x[maxn];

64     

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

66         sum=0;

67         build(0,n-1,1);

68         

69         for(int i = 0;i < n;i++) {

70             scanf("%d",&x[i]);

71             sum+=query(x[i],n-1,0,n-1,1);

72             Updata(x[i],0,n-1,1);

73         }

74         int ret=sum;

75         for(int i=0;i<n;i++) {

76             sum+=n-x[i]-x[i]-1;

77             ret=min(ret,sum);

78         }

79         

80         printf("%d\n",ret);

81     }

82     

83     return 0;

84 }