hdu3308LCIS

最新推荐文章于 2018-10-18 20:43:00 发布

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本文介绍了一个典型的数据结构问题——最长连续递增子序列（LCIS）的求解方法。通过构建特殊的区间树来高效地处理更新和查询操作，实现对序列的修改及查询LCIS长度的功能。

http://acm.hdu.edu.cn/showproblem.php?pid=3308

LCIS

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6448 Accepted Submission(s): 2799

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

分析：典型的区间合并题目，是入门必做的区间合并类……

在构建的时候，需要同时构造包括左端点的区间lx、包括右端点的区间rx以及该区间的总的长度mx……在更新的时候需要判断在端点处的值是否符合递增……

当符合的时候，把左孩子和右孩子的LCIS整合起来，更新mx；当不符合时mx取lx和rx中的最大值。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <map>
#include <cmath>
#include <vector>
using namespace std;

#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1

#define inf 0x3f3f3f3f
const int maxn= 100000+7;
typedef long long ll;
int lx[maxn<<2],mx[maxn<<2],rx[maxn<<2];
int a[maxn];
void pushup(int rt,int l,int r) {
    int len = r-l+1;
    int m = r+l>>1;
    lx[rt] = lx[rt<<1];
    rx[rt] = rx[rt<<1|1];
    if((lx[rt] == (len-(len>>1))) && (a[m] < a[m+1]))lx[rt] += lx[rt<<1|1];
    if((rx[rt] == (len>>1)) && (a[m] < a[m+1]))rx[rt] += rx[rt<<1];
    mx[rt] = max(max(mx[rt<<1],mx[rt<<1|1]),(a[m] < a[m+1])?rx[rt<<1]+lx[rt<<1|1]:0);
}
void build(int l,int r,int rt) {
    if(l == r) {
        scanf("%d",&a[l]);
        lx[rt] = mx[rt] = rx[rt] = 1;
        return;
    }
    int mid = l+r>>1;
    build(lson);
    build(rson);
    pushup(rt,l,r);
}

void update(int pos,int w,int l,int r,int rt) {
    if(l == r) {
        a[l] = w;
        lx[rt] = rx[rt] = mx[rt] = 1;
        return;
    }
    int mid = l+r>>1;
    if(pos <= mid)update(pos,w,lson);
    else update(pos,w,rson);
    pushup(rt,l,r);
}
int query(int L,int R,int l,int r,int rt) {
    if(L <= l && r <= R)
        return mx[rt];
    int mid = l+r>>1;
    if(R <= mid)return query(L,R,lson);
    else if(L > mid)return query(L,R,rson);
    else {
        int ltmp = query(L,mid,lson);
        int rtmp = query(mid+1,R,rson);
        int sum = 0;
        if(a[mid+1] > a[mid]) {
            sum = min(mid-L+1,rx[rt<<1])+min(R-mid,lx[rt<<1|1]);
        }
        return max(sum,max(ltmp,rtmp));
    }
}
int main() {
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int n,m,b,c,t;
    char ss[10];
    scanf("%d",&t);
    while(t--) {
        scanf("%d%d",&n,&m);
        build(0,n-1,1);
        for(int i = 1; i <= m; i++) {
            scanf("%s%d%d",ss,&b,&c);
            if(ss[0] == 'U') {
                update(b,c,0,n-1,1);
            } else {
                printf("%d\n",query(b,c,0,n-1,1));
            }
        }
    }
    return 0;
}