hdu LCIS（线段树）

LCIS

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1713    Accepted Submission(s): 748

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

   
   
   
   

    
    
    
    1
1
4
2
3
1
2
5
   
   
   
   

 

Author

shǎ崽

 

Source

HDOJ Monthly Contest – 2010.02.06

 

Recommend

wxl

这道题能很能体现线段树的通用性，通过区间节点保存的信息，大大优化了时间

从节点里保存的很多信息可以看出来，线段树题的关键就在于储存哪些信息，和如何利用这些信息

需要缜密的思考tree[n].num,tree[n].tagl,tree[n].tagr,这些值如何更新

这道题可能很神奇的1Y，可惜我写错了一个地方，没体会到这种乐趣。。。

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

struct Node{
    int l,r,num;
    int first,end;
    int tagl,tagr;
}tree[1000000];

void Build(int n,int x,int y){
    tree[n].l = x;
    tree[n].r = y;
    int mid = (x + y) / 2;
    if(x == y){
        int a;
        scanf("%d",&a);
        tree[n].first = a;//区间首元素的值
        tree[n].end = a;//区间尾元素。。
        tree[n].num = 1;//区间内最长序列
        tree[n].tagl = 1;//从左起最长上升序列
        tree[n].tagr = 1;//从右起。。。。。。
        return;
    }
    Build(2*n,x,mid);
    Build(2*n+1,mid+1,y);
    tree[n].first = tree[2*n].first;
    tree[n].end = tree[2*n+1].end;
    //如果左儿子整个区间连续，且末元素<右儿子区间首元素
    if(tree[2*n].num == (tree[2*n].r - tree[2*n].l + 1) && tree[2*n].end < tree[2*n+1].first)
        tree[n].tagl = tree[2*n].num + tree[2*n+1].tagl;
    else
        tree[n].tagl = tree[2*n].tagl;
    //tagr同理
    if(tree[2*n+1].num == (tree[2*n+1].r - tree[2*n+1].l + 1) && tree[2*n].end < tree[2*n+1].first)
        tree[n].tagr = tree[2*n+1].num + tree[2*n].tagr;
    else
        tree[n].tagr = tree[2*n+1].tagr;
    //求tree[n].num
    if(tree[2*n].end < tree[2*n+1].first)
        tree[n].num = max(tree[2*n].tagr + tree[2*n+1].tagl,max(tree[2*n].num,tree[2*n+1].num));
    else
        tree[n].num = max(tree[2*n].num,tree[2*n+1].num);
}

void Modify(int n,int x,int y){
    int l = tree[n].l;
    int r = tree[n].r;
    int mid = (l + r) / 2;
    if(l == r){
        tree[n].first = y;
        tree[n].end = y;
        return;
    }
    if(x <= mid)    Modify(2*n,x,y);
    else            Modify(2*n+1,x,y);
    //更新节点中的所有值，和Build中一样
    tree[n].first = tree[2*n].first;
    tree[n].end = tree[2*n+1].end;
    if(tree[2*n].num == (tree[2*n].r - tree[2*n].l + 1) && tree[2*n].end < tree[2*n+1].first)
        tree[n].tagl = tree[2*n].num + tree[2*n+1].tagl;
    else
        tree[n].tagl = tree[2*n].tagl;
    if(tree[2*n+1].num == (tree[2*n+1].r - tree[2*n+1].l + 1) && tree[2*n].end < tree[2*n+1].first)
        tree[n].tagr = tree[2*n+1].num + tree[2*n].tagr;
    else
        tree[n].tagr = tree[2*n+1].tagr;
    if(tree[2*n].end < tree[2*n+1].first)
        tree[n].num = max(tree[2*n].tagr + tree[2*n+1].tagl,max(tree[2*n].num,tree[2*n+1].num));
    else
        tree[n].num = max(tree[2*n].num,tree[2*n+1].num);
}

int Searchl(int n,int x,int y){
    int l = tree[n].l;
    int r = tree[n].r;
    int mid = (l + r) / 2;
    if(l == x && r == y){
        return tree[n].tagl;
    }
    if(x <= mid && y <= mid)
        return Searchl(2*n,x,y);
    else if(x > mid && y > mid)
        return Searchl(2*n+1,x,y);
    else if(Searchl(2*n,x,mid) == (mid - x + 1) && tree[2*n].end < tree[2*n+1].first)
        return (mid - x + 1) +  Searchl(2*n+1,mid+1,y);
    else
        return Searchl(2*n,x,mid);
}

int Searchr(int n,int x,int y){
    int l = tree[n].l;
    int r = tree[n].r;
    int mid = (l + r) / 2;
    if(l == x && r == y){
        return tree[n].tagr;
    }
    if(x <= mid && y <= mid)
        return Searchr(2*n,x,y);
    else if(x > mid && y > mid)
        return Searchr(2*n+1,x,y);
    else if(Searchr(2*n+1,mid+1,y) == (y - (mid + 1) + 1) && tree[2*n].end < tree[2*n+1].first)
        return Searchr(2*n,x,mid) +  (y - mid);
    else
        return Searchr(2*n+1,mid+1,y);
}

int Find(int n,int x,int y){
    int l = tree[n].l;
    int r = tree[n].r;
    int mid = (l + r) / 2;
    if(l == x && r == y){
        return tree[n].num;
    }
    int ans;
    if(x <= mid && y <= mid)    ans = Find(2*n,x,min(y,mid));
    else if(y > mid && x > mid)   ans = Find(2*n+1,max(x,mid+1),y);
    else if(tree[2*n].end < tree[2*n+1].first)
        ans = max(Searchr(2*n,x,mid) + Searchl(2*n+1,mid+1,y),max(Find(2*n,x,min(y,mid)),Find(2*n+1,max(x,mid+1),y)));
    else
        ans = max(Find(2*n,x,min(y,mid)),Find(2*n+1,max(x,mid+1),y));
    return ans;
}

int main(){
    int t,m,n;
    int i;
    char str[10];
    int a,b;

    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        Build(1,0,n-1);
        for(i = 0;i < m;i++){
            scanf("%s%d%d",str,&a,&b);//及时只有一个字符，用%s存而不用%c存，避免读取时空格、回车这些干扰
            if(str[0] == 'Q')    printf("%d\n",Find(1,a,b));
            else    Modify(1,a,b);
        }
    }
    return 0;
}