hdu3308 LCIS(简单线段树)

LCIS

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2988    Accepted Submission(s): 1307

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

 

Author

shǎ崽

题目大意：给一串数字，1种操作，将第a个数改成b；一种查询，查询区间[a,b]的LICS。

题目分析：跟这种题一个类型，同时维护端点左右连续值就可以了。详情请见代码：

#include <iostream>#include<cstdio>#include<cstring>using namespace std;const int N = 100005;struct node{    int lc,rc;//区间左右LCIS    int lv,rv;//区间左右端点值    int ans;//区间LCIS}tree[N<<2];int m,n;int Max(int a,int b){    return a > b?a:b;}int Min(int a,int b){    return a > b?b:a;}void pushup(int num,int s,int e){    int ls = num<<1;    int rs = num<<1|1;    int mid = (s + e)>>1;    tree[num].lv = tree[ls].lv;    tree[num].rv = tree[rs].rv;    tree[num].lc = tree[ls].lc;    if(tree[ls].ans == mid - s + 1 && tree[ls].rv < tree[rs].lv)        tree[num].lc += tree[rs].lc;    tree[num].rc = tree[rs].rc;    if(tree[rs].ans == e - mid && tree[rs].lv > tree[ls].rv)        tree[num].rc += tree[ls].rc;    int cmp = Max(tree[ls].rc,tree[rs].lc);    if(tree[ls].rv < tree[rs].lv)        cmp = tree[ls].rc + tree[rs].lc;    tree[num].ans = Max(tree[ls].ans,Max(tree[rs].ans,cmp));}void build(int num,int s,int e){    if(s == e)    {        scanf("%d",&tree[num].lv);        tree[num].rv = tree[num].lv;        tree[num].lc = tree[num].rc = 1;        tree[num].ans = 1;        return;    }    int mid = (s + e)>>1;    build(num<<1,s,mid);    build(num<<1|1,mid + 1,e);    pushup(num,s,e);}void update(int num,int s,int e,int pos,int val){    if(s == e)    {        tree[num].lv = tree[num].rv = val;        return;    }    int mid = (s + e)>>1;    if(pos <= mid)        update(num<<1,s,mid,pos,val);    else        update(num<<1|1,mid + 1,e,pos,val);    pushup(num,s,e);}int query(int num,int s,int e,int l,int r){    if(s == l && r == e)    {        return tree[num].ans;    }    int mid = (s + e)>>1;    if(r <= mid)        return query(num<<1,s,mid,l,r);    else    {        if(l > mid)            return query(num<<1|1,mid + 1,e,l,r);        else        {            int ta = Min(tree[num<<1].rc,mid - l + 1);            int tb = Min(tree[num<<1|1].lc,r - mid);            int middle = Max(ta,tb);            if(tree[num<<1].rv < tree[num<<1|1].lv)                middle = ta + tb;            return Max(query(num<<1,s,mid,l,mid),Max(query(num<<1|1,mid + 1,e,mid + 1,r),middle));        }    }}int main(){    int t;    char op[3];    int a,b;    scanf("%d",&t);    while(t --)    {        scanf("%d%d",&n,&m);        build(1,1,n);        while(m --)        {            scanf("%s",op);            scanf("%d%d",&a,&b);            if(op[0] == 'Q')                printf("%d\n",query(1,1,n,a + 1,b + 1));            else                update(1,1,n,a + 1,b);        }    }    return 0;}//343MS5376K