hdu3308 LCIS

最新推荐文章于 2021-03-30 11:33:17 发布

magicnumber

最新推荐文章于 2021-03-30 11:33:17 发布

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本文探讨了LCIS问题的解决策略，并结合区间合并、维护最左元素LCIS、最右元素LCIS及整体LCIS的方法进行了解析。通过实例输入输出展示了算法的应用过程，包括更新操作和查询最长连续递增子序列的长度。

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LCIS

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3336 Accepted Submission(s): 1476

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<=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).

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

Author

shǎ崽

Source

HDOJ Monthly Contest – 2010.02.06

区间合并，

维护一个区间的包含最左的元素的LCIS，包含最右元素的LCIS，以及整个区间的LCIS，然后更新的时候就更新这三个值就行了。

代码

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

typedef struct
{
    int l,r,val,lval,rval;
}Tree;

Tree tree[3000000];
int a[500005];

void PushUp(int t)
{
  //  printf("%d %d %d %d !!!!\n",tree[2*t+1].r,tree[2*t+2].l,a[tree[2*t+1].r],a[tree[2*t+2].l]);
    tree[t].val=max(tree[2*t+1].val,tree[2*t+2].val);
    if (a[tree[2*t+1].r]>=a[tree[2*t+2].l])
    {
        tree[t].lval=tree[2*t+1].lval;
        tree[t].rval=tree[2*t+2].rval;
    }
    else
    {
        tree[t].val=max(tree[t].val,tree[2*t+1].rval+tree[2*t+2].lval);
        if (tree[2*t+1].lval==tree[2*t+1].r-tree[2*t+1].l+1) tree[t].lval=tree[2*t+1].lval+tree[2*t+2].lval;
        else tree[t].lval=tree[2*t+1].lval;
        if (tree[2*t+2].rval==tree[2*t+2].r-tree[2*t+2].l+1) tree[t].rval=tree[2*t+1].rval+tree[2*t+2].rval;
        else tree[t].rval=tree[2*t+2].rval;
        tree[t].val=max(max(tree[t].val,tree[t].lval),tree[t].rval);
    }
  //  printf("%d %d %d %d %d\n",tree[t].l,tree[t].r,tree[t].val,tree[t].lval,tree[t].rval);
}

void Build(int t,int l,int r)
{
    tree[t].l=l;
    tree[t].r=r;
    if (l==r)
    {
        scanf("%d",&a[l]);
        tree[t].val=a[l];
        tree[t].lval=tree[t].rval=tree[t].val=1;
        return;
    }
    int mid=(l+r)>>1;
    Build(2*t+1,l,mid);
    Build(2*t+2,mid+1,r);
    PushUp(t);
}

int Query(int t,int l,int r)
{
    if (tree[t].l==l && tree[t].r==r)
    {
        return tree[t].val;
    }
    int mid=(tree[t].l+tree[t].r)>>1;
    int cnt=0;
    if (l<=mid) cnt=max(cnt,Query(2*t+1,l,min(r,mid)));
    if (r>mid) cnt=max(cnt,Query(2*t+2,max(l,mid+1),r));
    if (l>mid || r<=mid || a[tree[2*t+1].r]>=a[tree[2*t+2].l]) return cnt;
    int tmp=min(tree[2*t+1].rval,mid-l+1)+min(tree[2*t+2].lval,r-mid);
    return max(tmp,cnt);
}

void Update(int t,int x,int val)
{
    if (tree[t].l==tree[t].r)
    {
        a[tree[t].l]=val;
        return;
    }
    int mid=(tree[t].l+tree[t].r)>>1;
    if (x<=mid) Update(2*t+1,x,val);
    else Update(2*t+2,x,val);
    PushUp(t);
}

int main()
{
    int i,j,n,T,m,x,y;
    char str[5];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        Build(0,0,n-1);
        while(m--)
        {
            scanf("%s%d%d",str,&x,&y);
            if (str[0]=='Q')
            {
                printf("%d\n",Query(0,x,y));
            }
            else
            {
                Update(0,x,y);
            }
        }
    }
    return 0;
}