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

[b]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].
[/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).

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。

#include <iostream>
#include <cstdio>
using namespace std;
const int maxn = 100010;
struct node{
int lt,rt,lx,rx,mx,wth;
}tree[maxn<<2];
int d[maxn];
void fix(int v,int mid){
tree[v].lx = tree[v<<1].lx;
tree[v].rx = tree[v<<1|1].rx;
if(d[mid] < d[mid+1]){
tree[v].lx += tree[v].lx == tree[v<<1].wth?tree[v<<1|1].lx:0;
tree[v].rx += tree[v].rx == tree[v<<1|1].wth?tree[v<<1].rx:0;
tree[v].mx = max(tree[v<<1].mx,tree[v<<1|1].mx);
tree[v].mx = max(tree[v].mx,tree[v<<1].rx+tree[v<<1|1].lx);
}else tree[v].mx = max(tree[v<<1].mx,tree[v<<1|1].mx);
}
void build(int lt,int rt,int v){
tree[v].lt = lt;
tree[v].rt = rt;
tree[v].wth = rt-lt+1;
if(lt == rt){tree[v].lx = tree[v].rx = tree[v].mx = 1;return;}
int mid = (lt+rt)>>1;
build(lt,mid,v<<1);
build(mid+1,rt,v<<1|1);
fix(v,mid);
}
void update(int u,int val,int v){
int mid = (tree[v].lt+tree[v].rt)>>1;
if(tree[v].lt == tree[v].rt) {d[tree[v].lt] = val;return;}
if(u <= mid) update(u,val,v<<1);
else update(u,val,v<<1|1);
fix(v,mid);
}
int query(int lt,int rt,int v){
if(tree[v].lt == lt && tree[v].rt == rt) return tree[v].mx;
int mid = (tree[v].lt+tree[v].rt)>>1;
if(rt <= mid) return query(lt,rt,v<<1);
else if(lt > mid) return query(lt,rt,v<<1|1);
else{
int temp = max(query(lt,mid,v<<1),query(mid+1,rt,v<<1|1));
if(d[mid] < d[mid+1]){
temp = max(min(mid-lt+1,tree[v<<1].rx)+min(rt-mid,tree[v<<1|1].lx),temp);
//看看合并是否可以使取值更优
}
return temp;
}
}
int main(){
int ks,n,m,i,x,y;
char op[5];
scanf("%d",&ks);
while(ks--){
scanf("%d %d",&n,&m);
for(i = 1; i <= n; i++)
scanf("%d",d+i);
build(0,n,1);
for(i = 0; i < m; i++){
scanf("%s%d%d",op,&x,&y);
if(op[0] == 'Q'){
printf("%d\n",query(x+1,y+1,1));
}else update(x+1,y,1);
}
}
return 0;
}

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