【LOJ】#2056. 「TJOI / HEOI2016」序列

题解

这个我们处理出来每一位能变化到的最大值和最小值，包括自身

然后我们发现
\(f[i] = max(f[i],f[j] + 1) (mx[j] <= a[i] && a[j] <= mi[i])\)
喜闻乐见的三维偏序转移法
还写树套树？？？
直接CDQ分治就好啦

为啥他们的代码就1.xK？？？？？？
我就3.1K

代码

#include <bits/stdc++.h>
#define enter putchar('\n')
#define space putchar(' ')
#define pii pair<int,int>
#define fi first
#define se second
#define MAXN 100005
#define pb push_back
#define mp make_pair
#define eps 1e-8
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 + c - '0';
        c = getchar();
    }
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) out(x / 10);
    putchar('0' + x % 10);
}
vector<int> tr[MAXN * 4];
int mx[MAXN],mi[MAXN],a[MAXN],N,M,f[MAXN],p[MAXN],tmp[MAXN],g[MAXN];
int BIT[MAXN * 2];
int lowbit(int u) {return u & -u;}
void Insert(int u,int v) {
    while(u <= 100000) {
        BIT[u] = max(BIT[u],v);
        u += lowbit(u);
    }
}
int Query(int u) {
    int res = 0;
    while(u > 0) {
        res = max(res,BIT[u]);
        u -= lowbit(u);
    }
    return res;
}
void Clear(int u) {
    while(u <= 100000) {
        BIT[u] = 0;
        u += lowbit(u);
    }
}
bool cmp(int c,int d) {
    return a[c] < a[d];
}
void build(int u,int l,int r) {
    if(l == r) {tr[u].pb(l);return;}
    int mid = (l + r) >> 1;
    build(u << 1,l,mid);build(u << 1 | 1,mid + 1,r);
    int s1 = 0,s2 = 0;
    while(s1 < tr[u << 1].size() || s2 < tr[u << 1 | 1].size()) {
        if(s2 >= tr[u << 1 | 1].size()) tr[u].pb(tr[u << 1][s1++]);
        else if(s1 >= tr[u << 1].size() || a[tr[u << 1 | 1][s2]] < a[tr[u << 1][s1]]) tr[u].pb(tr[u << 1 | 1][s2++]);
        else tr[u].pb(tr[u << 1][s1++]);
    }
}
void Solve(int u,int l,int r) {
    if(l == r) {return ;}
    int mid = (l + r) >> 1;
    Solve(u << 1,l,mid);
    int s1 = l,s2 = 0,t;
    while(s1 <= mid || s2 < tr[u << 1 | 1].size()) {
        if(s2 >= tr[u << 1 | 1].size()) break;
        if(s1 > mid || a[tr[u << 1 | 1][s2]] < mx[p[s1]]) {
            t = tr[u << 1 | 1][s2];++s2;
            f[t] = max(f[t],Query(mi[t]) + 1);
        }
        else {
            Insert(a[p[s1]],f[p[s1]]);
            ++s1;
        }
    }
    for(int i = l ; i <= mid ; ++i) Clear(a[i]);
    Solve(u << 1 | 1,mid + 1,r);
    s1 = l,s2 = mid + 1,t = l;
    while(s1 <= mid || s2 <= r) {
        if(s2 > r) tmp[t++] = p[s1++];
        else if(s1 > mid || mx[p[s1]] > mx[p[s2]]) tmp[t++] = p[s2++];
        else tmp[t++] = p[s1++];
    }
    for(int i = l ; i <= r ; ++i) p[i] = tmp[i];
    return;
}
void Init() {
    read(N);read(M);
    for(int i = 1 ; i <= N ; ++i) {read(a[i]);mi[i] = mx[i] = a[i];p[i] = i;f[i] = 1;}
    int u,v;
    for(int i = 1 ; i <= M ; ++i) {
        read(u);read(v);
        mi[u] = min(mi[u],v);
        mx[u] = max(mx[u],v);
    }
    build(1,1,N);
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve(1,1,N);
    int ans = 0;
    for(int i = 1 ; i <= N ; ++i) ans = max(ans,f[i]);
    out(ans);enter;
    return 0;
}