HDU3564 --- Another LIS （线段树维护最值问题）

Another LIS

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1244    Accepted Submission(s): 431

Problem Description

There is a sequence firstly empty. We begin to add number from 1 to N to the sequence, and every time we just add a single number to the sequence at a specific position. Now, we want to know length of the LIS (Longest Increasing Subsequence) after every time's add.

 

 

Input

An integer T (T <= 10), indicating there are T test cases.
For every test case, an integer N (1 <= N <= 100000) comes first, then there are N numbers, the k-th number Xk means that we add number k at position Xk (0 <= Xk <= k-1).See hint for more details.

 

 

Output

For the k-th test case, first output "Case #k:" in a separate line, then followed N lines indicating the answer. Output a blank line after every test case.

 

 

Sample Input

1

3

0 0 2

 

 

Sample Output

Case #1:

1

1

2

Hint

In the sample, we add three numbers to the sequence, and form three sequences. a. 1 b. 2 1 c. 2 1 3

 

题意：一个空序列，第k次在xk处 插入 k，并输出 序列当前的LIS。

首先 倒序 用线段树来确定每个元素 最终的位置。然后 求n次LIS，当然要用线段树维护了，，或者二分等等也可以。 nlogn

 1 #include <cstdio>

 2 #include <string>

 3 #include <cstdlib>

 4 #include <cstring>

 5 #include <algorithm>

 6 using namespace std;

 7 const int maxn = 1e5+10;

 8 int seg[maxn<<2],a[maxn];            // seg表示该区间 位置剩余量

 9 void build (int l,int r,int pos)

10 {

11     if (l == r)

12     {

13         seg[pos] = 1;

14         return;

15     }

16     int mid = (l + r) >> 1;

17     build(l,mid,pos<<1);

18     build(mid+1,r,pos<<1|1);

19     seg[pos] = seg[pos<<1] + seg[pos<<1|1];

20 }

21 int ans[maxn];                            //ans[i]表示 元素i的位置为ans[i]

22 void Insert(int l,int r,int pos,int x,int k)

23 {

24     if (l == r)

25     {

26         ans[k] = l;

27         seg[pos] = 0;

28         return ;

29     }

30     int mid = (l + r) >> 1;

31     if (seg[pos<<1] >= x)                            //左孩子的区间 位置剩余量大于x时，递归进入左孩子，否则右孩子同时 x-seg[pos<<1]

32         Insert(l,mid,pos<<1,x,k);

33     else

34         Insert(mid+1,r,pos<<1|1,x - seg[pos<<1],k);

35     seg[pos] = seg[pos<<1] + seg[pos<<1|1];

36 }

37 void update(int l,int r,int pos,int x,int v)

38 {

39     if (l == r)

40     {

41         seg[pos] = v;

42         return;

43     }

44     int mid = (l + r) >> 1;

45     if (x <= mid)

46         update(l,mid,pos<<1,x,v);

47     else

48         update(mid+1,r,pos<<1|1,x,v);

49     seg[pos] = max(seg[pos<<1],seg[pos<<1|1]);

50 }

51 int query(int l,int r,int pos,int ua,int ub)

52 {

53     if (ua <= l && ub >= r)

54     {

55         return seg[pos];

56     }

57     int mid = (l + r) >> 1;

58     int t1 = 0,t2 = 0;

59     if (ua <= mid)

60         t1 = query(l,mid,pos<<1,ua,ub);

61     if (ub > mid)

62         t2 = query(mid+1,r,pos<<1|1,ua,ub);

63     return max(t1,t2);

64 }

65 

66 int main(void)

67 {

68     #ifndef ONLINE_JUDGE

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

70     #endif

71     int t,cas = 1;

72     scanf ("%d",&t);

73     while (t--)

74     {

75         int n;

76         scanf ("%d",&n);

77         for (int i = 1; i <= n; i++)

78             scanf ("%d",a+i);

79         build (1,n,1);

80         for (int i = n; i >= 1; i--)

81             Insert(1,n,1,a[i] + 1,i);               //倒序 确定每个元素 最终的位置保存在ans里

82         memset(seg,0,sizeof(seg));

83         printf("Case #%d:\n",cas++);

84         for (int i = 1; i <= n; i++)

85         {

86             int t1 = query(1,n,1,1,n);

87             int t2 = query(1,n,1,1,ans[i]);

88             update(1,n,1,ans[i],t2+1);

89             if (t2 < t1)                        //输出当前区间的LIS

90                 printf("%d\n",t1);

91             else

92                 printf("%d\n",1+t2);

93         }

94         printf("\n");

95     }

96     return 0;

97 }