HDU 3308 LCIS

 

LCIS

 

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on  HDU. Original ID: 3308
64-bit integer IO format: %I64d      Java class name: Main

 

Given n integers.
You have two operations:
U A B: replace the Ath number by B. (index counting from 0)
Q A B: output the length of the longest consecutive increasing subsequence (LCIS) in [a, b].

 

Input

T in the first line, indicating the case number.
Each case starts with two integers n , m(0<n,m<=10 ⁵).
The next line has n integers(0<=val<=10 ⁵).
The next m lines each has an operation:
U A B(0<=A,n , 0<=B=10 ⁵)
OR
Q A B(0<=A<=B< n).

 

Output

For each Q, output the answer.

 

Sample Input

1

10 10

7 7 3 3 5 9 9 8 1 8 

Q 6 6

U 3 4

Q 0 1

Q 0 5

Q 4 7

Q 3 5

Q 0 2

Q 4 6

U 6 10

Q 0 9

 

Sample Output

1

1

4

2

3

1

2

5

解题思路：线段树的区间合并。哎。。。写了三四遍才过的！！！！！！！！int lt,rt,lx,rx,mx,wth;依次为左边界，右边界，左上升区间长度，右上升区间长度，lt rt中的最大上升区间长度。

 1 #include <iostream>

 2 #include <cstdio>

 3 using namespace std;

 4 const int maxn = 100010;

 5 struct node{

 6     int lt,rt,lx,rx,mx,wth;

 7 }tree[maxn<<2];

 8 int d[maxn];

 9 void fix(int v,int mid){

10     tree[v].lx = tree[v<<1].lx;

11     tree[v].rx = tree[v<<1|1].rx;

12     if(d[mid] < d[mid+1]){

13         tree[v].lx += tree[v].lx == tree[v<<1].wth?tree[v<<1|1].lx:0;

14         tree[v].rx += tree[v].rx == tree[v<<1|1].wth?tree[v<<1].rx:0;

15         tree[v].mx = max(tree[v<<1].mx,tree[v<<1|1].mx);

16         tree[v].mx = max(tree[v].mx,tree[v<<1].rx+tree[v<<1|1].lx);

17     }else tree[v].mx = max(tree[v<<1].mx,tree[v<<1|1].mx);

18 }

19 void build(int lt,int rt,int v){

20     tree[v].lt = lt;

21     tree[v].rt = rt;

22     tree[v].wth = rt-lt+1;

23     if(lt == rt){tree[v].lx = tree[v].rx = tree[v].mx = 1;return;}

24     int mid = (lt+rt)>>1;

25     build(lt,mid,v<<1);

26     build(mid+1,rt,v<<1|1);

27     fix(v,mid);

28 }

29 void update(int u,int val,int v){

30     int mid = (tree[v].lt+tree[v].rt)>>1;

31     if(tree[v].lt == tree[v].rt) {d[tree[v].lt] = val;return;}

32     if(u <= mid) update(u,val,v<<1);

33     else update(u,val,v<<1|1);

34     fix(v,mid);

35 }

36 int query(int lt,int rt,int v){

37     if(tree[v].lt == lt && tree[v].rt == rt) return tree[v].mx;

38     int mid = (tree[v].lt+tree[v].rt)>>1;

39     if(rt <= mid) return query(lt,rt,v<<1);

40     else if(lt > mid) return query(lt,rt,v<<1|1);

41     else{

42         int temp = max(query(lt,mid,v<<1),query(mid+1,rt,v<<1|1));

43         if(d[mid] < d[mid+1]){

44             temp = max(min(mid-lt+1,tree[v<<1].rx)+min(rt-mid,tree[v<<1|1].lx),temp);

45             //看看合并是否可以使取值更优

46         }

47         return temp;

48     }

49 }

50 int main(){

51     int ks,n,m,i,x,y;

52     char op[5];

53     scanf("%d",&ks);

54     while(ks--){

55         scanf("%d %d",&n,&m);

56         for(i = 1; i <= n; i++)

57             scanf("%d",d+i);

58             build(0,n,1);

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

60             scanf("%s%d%d",op,&x,&y);

61             if(op[0] == 'Q'){

62                 printf("%d\n",query(x+1,y+1,1));

63             }else update(x+1,y,1);

64         }

65     }

66     return 0;

67 }

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