LCIS
Time Limit: 2000ms
Memory Limit: 32768KB
This problem will be judged on HDU. Original ID: 330864-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].
View Code
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<=105).
The next line has n integers(0<=val<=105).
The next m lines each has an operation:
U A B(0<=A,n , 0<=B=105)
OR
Q A B(0<=A<=B< n).
Each case starts with two integers n , m(0<n,m<=105).
The next line has n integers(0<=val<=105).
The next m lines each has an operation:
U A B(0<=A,n , 0<=B=105)
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 ",query(x+1,y+1,1)); 63 }else update(x+1,y,1); 64 } 65 } 66 return 0; 67 }