Hdu 3308 LCIS
LCIS
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1611 Accepted Submission(s): 717
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].
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).
lMax = (左儿子的lMax == 左儿子的len) ? 左儿子的len + 右儿子的lMax : 左儿子的lMax; rMax = (右儿子的rMax == 右儿子的len) ? 右儿子的len + 左儿子的rMax : 右儿子的rMax; Max = MAX(左儿子的rMax + 右儿子的lMax, 左儿子的Max, 右儿子的Max, lMax, rMax);2. 左儿子最右边的值 >= 右儿子最左边的值 lMax = 左儿子的lMax; rMax = 右儿子的rMax; Max = MAX(左儿子的Max, 右儿子的Max);
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
#define N 100003
using namespace std;
struct node
{
int lmax,rmax,max;
int lva,rva;
int len;
};
node st[N<<3];
void up(node &k,node &ls,node &rs)
{
k.lva=ls.lva;
k.rva=rs.rva;
//k.len=ls.len+rs.len;
if(ls.rva<rs.lva)//开始这里的判断写反了、、、郁闷
{
k.lmax=(ls.len==ls.lmax?ls.lmax+rs.lmax:ls.lmax);
k.rmax=(rs.len==rs.rmax?ls.rmax+rs.rmax:rs.rmax);
k.max=max(max(k.lmax,k.rmax),ls.rmax+rs.lmax);
k.max=max(max(ls.max,rs.max),k.max);
}
else
{
k.lmax=ls.lmax;
k.rmax=rs.rmax;
k.max=max(ls.max,rs.max);
}
}
void build(int l,int r,int k)
{
st[k].len=r-l+1;
if(l==r)
{
scanf("%d",&st[k].lva);//printf("l=%d %d\n",l,st[k].lva);
st[k].rva=st[k].lva;
st[k].max=st[k].lmax=st[k].rmax=1;
return ;
}
int m=(l+r)>>1;
build(lson);
build(rson);
int t=k<<1;
up(st[k],st[t],st[t+1]);
}
int va;
void updata(int &index,int l,int r,int k)
{
if(l==r)
{
st[k].lva=st[k].rva=va;
//printf("%d %d %d\n",l,r,st[k].lva);
return ;
}
int m=(l+r)>>1;
if(index<=m) updata(index,lson);
else updata(index,rson);
int t=k<<1;
up(st[k],st[t],st[t+1]);
}
node Query(int &L,int &R,int l,int r,int k)
{ // printf("%d %d\n",l,r);
if(L<=l&&R>=r)
return st[k];
int m=(l+r)>>1;
node r1,r2;
r1.len=r2.len=0;
if(L<=m) r1=Query(L,R,lson);
if(R>m) r2=Query(L,R,rson);
if(r1.len&&r2.len)
{
node temp;
//printf("k=%d %d\n",r1.max,r2.max);
up(temp,r1,r2);
temp.len=r1.len+r2.len;
return temp;
}
if(r1.len) return r1;
if(r2.len) return r2;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,m,x,y;
node te;
char c;
scanf("%d%d",&n,&m);
n--;
build(0,n,1);getchar();//坑爹的每组数据后面多了个空格、、、
// printf("%d ",st[1].max);
while(m--)
{
getchar();
scanf("%c%d%d",&c,&x,&y);
//printf("%c",c);
if(c=='Q')
{
te=Query(x,y,0,n,1);
printf("%d\n",te.max);
}
else
va=y,updata(x,0,n,1);
}
}
return 0;
}