HDOJ 3308 LCIS （线段树之区间合并）

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LCIS

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7349 Accepted Submission(s): 3110

Problem Description

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<=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

110 107 7 3 3 5 9 9 8 1 8 Q 6 6U 3 4Q 0 1Q 0 5Q 4 7Q 3 5Q 0 2Q 4 6U 6 10Q 0 9

Sample Output

11423125

#include <bits/stdc++.h>using namespace std;#define mst(a,b) memset((a),(b),sizeof(a))#define f(i,a,b) for(int i=(a);i<(b);++i)const int maxn =10005;const int mod = 10005;const int INF = 1e9;#define ll long long#define m ((l+r)>>1)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define rush() int T;scanf("%d",&T);while(T--)int msum[maxn<<2]; //区间内最长连续递增子序列长度.  int lsum[maxn<<2]; //从第一个数开始算起 到a[m]的最大连续递增序列长度int rsum[maxn<<2]; //从a[m+1]开始算起 到最后一个数的最大连续递增序列长度int num[maxn];void pushup(int l,int r,int rt){    lsum[rt]=lsum[rt*2];    rsum[rt]=rsum[rt*2+1];    msum[rt]=max(msum[rt*2],msum[rt*2+1]);    int len=r-l+1;    if(num[m]<num[m+1])    {        if(lsum[rt]==len-(len/2))            lsum[rt]+=lsum[rt*2+1];        if(rsum[rt]==len/2)            rsum[rt]+=rsum[rt*2];        msum[rt]=max(msum[rt],lsum[rt*2+1]+rsum[rt*2]);    }}void build(int l,int r,int rt){    if(l==r)    {        msum[rt]=lsum[rt]=rsum[rt]=1;        return ;    }    build(lson);    build(rson);    pushup(l,r,rt) ;}void update(int p,int l,int r,int rt){    if(l==r)        return;    if(p<=m)        update(p,lson);    else        update(p,rson);    pushup(l,r,rt);}int query(int x,int y,int l,int r,int rt){    if(x<=l&&r<=y)    {        return msum[rt];    }    if(y<=m) return query(x,y,lson);    if(x>m)  return query(x,y,rson);    int a,b;    a=query(x,y,lson);    b=query(x,y,rson);    int ans;    ans=max(a,b);    if(num[m]<num[m+1]) //区间合并    {        int c;        c=min(rsum[rt*2],m-x+1)+min(lsum[rt*2+1],y-m);//防炸        ans=max(c,ans);    }    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,M;        mst(lsum,0),mst(rsum,0),mst(msum,0),mst(num,0);        scanf("%d%d",&n,&M);        f(i,1,n)        {            scanf("%d",&num[i]);        }        build(1,n,1);        while(M--)        {            char ch[2];            scanf("%s",ch);            int a,b;            scanf("%d%d",&a,&b);            if(ch[0]=='U')            {                a++;                num[a]=b;                update(a,1,n,1);            }            else            {                a++,b++;                printf("%d\n",query(a,b,1,n,1));            }        }    }    return 0;}

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