Codeforces 474E - Pillars

一眼看上去非常像最长不下降子序列。

然后比赛的时候对每个答案长度为k的序列，维护最后一个数的最大值和最小值。

当时不知道为什么认为从长度最长倒推至前面不会太长，于是心满意足地敲了个O(n^2)。结果T了。。。

正确的做法应该用线段树维护，搜起来就是log（n）,总的就是O(N*logN);

用非递归的方法写的

只用了77ms

#include <iostream>
#include <cstdio>
using namespace std;

#define ll long long

const int INF = ;

struct node {
    ll key;
    int pos;
} dtmin[INF], dtmax[INF];
ll n, d, t, x, M;
ll pre[INF];
int f[INF], ans[INF], namax[INF], namin[INF];

inline int Searchmax (int x, ll k) {
    while (dtmax[x].key - k >= d) {
        if (x > M) return dtmax[x].pos;
        if (dtmax[x <<  | ].key - k >= d) x = x <<  | ;
        else
            x = x << ;
    }
    return ;
}
inline int Searchmin (int x, ll k) {
    while (k - dtmin[x].key >= d) {
        if (x > M) return dtmin[x].pos;
        if (k - dtmin[x <<  | ].key >= d) x = x <<  | ;
        else
            x = x << ;
    }
    return ;
}

inline void modify (int x, ll k, int i) {
    if (dtmax[x + M].key < k) namax[x] = i, dtmax[x + M].key = k;
    for (int t = x + M; dtmax[t >> ].key < k && t > ; t >>= )
        dtmax[t >> ] = dtmax[t];

    if (dtmin[x + M].key > k) namin[x] = i, dtmin[x + M].key = k;
    for (int t = x + M; dtmin[t >> ].key > k && t > ; t >>= )
        dtmin[t >> ]  = dtmin[t];
}

int main() {
    ios::sync_with_stdio (false);
    cin >> n >> d;
    cin>>x;
    M = n;
    int tem = ;
    for (int k = M; k; tem++, k >>= ) ;
    M =  << tem;
    for (int i = ; i < (M << ) + ; i++) {
        dtmax[i].key =  , dtmin[i].key = (1LL << );
        dtmax[i].pos = i - M, dtmin[i].pos = i - M;
    }
    dtmax[].key = dtmin[].key = x;
    t = f[] = ;
    modify (t, x, );
    for (int i = ,k; i <= n; i++) {
        cin >> x;
        if (k = Searchmax (, x) )
            if (f[i] < k + )
                f[i] = k  + , pre[i] = namax[k];

        if ((k = Searchmin (, x)))
            if (f[i] < k + )
                f[i] = k  + , pre[i] = namin[k];

        if (f[i] > t) t = f[i];
        if (f[i] == )   f[i] = ;
        modify (f[i], x, i);
    }
    cout << t << endl;
    int p = -;
    for (int i = ; i <= n; i++)  if (f[i] == t) { p = i; break; }
    for (int i = t; i; i--)
        ans[i] = p, p = pre[p];
    for (int i = ; i <= t; i++)
        cout << ans[i] << ' ';
}