Codeforces 1029B. Creating the Contest 动态规划O(nlogn)解法 及 单调队列O(n)解法

题目链接：http://codeforces.com/problemset/problem/1029/B

题目大意：从数组a中选出一些数组成数组b，要求 b[i+1]<=b[i]*2 。

一开始想到的是O(n^2)的动态规划，但是超时了，下面是超时的代码。

#include <iostream>
using namespace std;

const int maxn = 200020;

int n, a[maxn], f[maxn], res = 0;

int main() {
    cin >> n;
    for (int i = 0; i < n; i ++) cin >> a[i];
    for (int i = 0; i < n; i ++) {
        f[i] = 1;
        for (int j = i-1; j >= 0 && a[i] <= 2*a[j]; j --) {
            f[i] = max(f[i], f[j]+1);
        }
        if (f[i] > res) res = f[i];
    }
    cout << res << endl;
    return 0;
}

然后想到的是：

• 二分找最小的满足a[i]<=a[j]*2的那个j ，O(logn)的时间复杂度
• 线段树求区间[j,i-1]的最大值，O(logn)的时间复杂度

再加上外层的循环，时间复杂度会降到O(n * logn)。

代码：

#include <iostream>
using namespace std;

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

const int maxn = 200020;

int n, a[maxn], MAX[maxn<<2];

void pushUp(int rt) {
    MAX[rt] = max(MAX[rt<<1] , MAX[rt<<1|1]);
}

void build(int l, int r, int rt) {
    if (l == r) {
        MAX[rt] = 0;
        return;
    }
    int m = (l+r) >> 1;
    build(lson);
    build(rson);
    pushUp(rt);
}

void update(int p, int val, int l, int r, int rt) {
    if (l == r) {
        MAX[rt] = val;
        return;
    }
    int m = (l + r) >> 1;
    if (p <= m) update(p, val, lson);
    else update(p, val, rson);
    pushUp(rt);
}

int query(int L, int R, int l, int r, int rt) {
    if (L <= l && r <= R) {
        return MAX[rt];
    }
    int m = (l + r) >> 1;
    int tmp = 0;
    if (L <= m) tmp = query(L, R, lson);
    if (R >= m+1) tmp = max(tmp, query(L, R, rson));
    return tmp;
}

int findL(int i) {
    return lower_bound(a, a+n+1, (a[i]+1)/2) - a;
}

int main() {
    cin >> n;
    build(1, n, 1);
    for (int i = 1; i <= n; i ++) cin >> a[i];
    for (int i = 1; i <= n; i ++) {
        int L = findL(i);
        int val;
        if (L >= i) val = 1;
        else {
            val = query(L, i-1, 1, n, 1) + 1;
        }
        update(i, val, 1, n, 1);
    }
    cout << query(1, n, 1, n, 1) << endl;
    return 0;
}

然后是O(n)的单调队列解法。

代码：

#include <iostream>
using namespace std;

const int maxn = 200020;

int n, a[maxn], MAX[maxn];
int st = 0, ed = 0, sum = 0;

int que[maxn];  // que存放id，id对应的最大长度是MAX[id]

void test() {
    cout << "[test]" << endl;
    for (int i = 1; i <= n; i ++) {
        cout << i << ": " << MAX[i] << endl;
    }
}

int main() {
    cin >> n;
    for (int i = 1; i <= n; i ++) cin >> a[i];
    int j = 1;
    for (int i = 1; i <= n; i ++) {
        while (j < i && !( a[i] <= 2 * a[j] )) {
            j ++;
        }
        while (st < ed && que[st] < j) st ++;
        if (st == ed) {
            MAX[i] = 1;
        } else {
            MAX[i] = MAX[ que[st] ] + 1;
        }
        sum = max(sum , MAX[i]);
        while (st < ed && MAX[ que[st] ] <= MAX[i]) {
            st ++;
        }
        que[ed++] = i;
    }
    // test();
    cout << sum << endl;
    return 0;
}