洛谷 P4093 [HEOI2016/TJOI2016]序列

题面

luogu

题解

\(Cdq分治+dp\)

\(mx[i],mn[i]\)分别表示第\(i\)位最大，最小能取到多少

那么有
\(j < i\)
\(mx[j] \le a[i]\)
\(a[j] \le mn[i]\)

然后就有了50分 \(O(n^2)\)的\(dp\)

上面那个东西是个三维偏序，
\(Cdq\)优化一下即可。

Code

50pts

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;
template<class T> inline void read(T &x) {
    x = 0; RG char c = getchar(); bool f = 0;
    while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
    while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
    x = f ? -x : x;
    return ;
}
template<class T> inline void write(T x) {
    if (!x) {putchar(48);return ;}
    if (x < 0) x = -x, putchar('-');
    int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
    for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}

const int N = 100010;
int a[N];
int mx[N], mn[N], f[N];

int main() {
    //freopen(".in", "r", stdin);
    //freopen(".out", "w", stdout);
    int n, m; read(n); read(m);
    for (int i = 1; i <= n; i++) read(a[i]), mx[i] = mn[i] = a[i], f[i] = 1;
    for (int i = 1; i <= m; i++) {
        int x, y; read(x), read(y);
        mx[x] = max(mx[x], y);
        mn[x] = min(mn[x], y);
    }
    int ans = 0;
    for (int i = 2; i <= n; i++)
        for (int j = 1; j < i; j++)
            if (a[i] >= mx[j] && mn[i] >= a[j])
                f[i] = max(f[i], f[j]+1), ans = max(ans, f[i]);
    printf("%d\n", ans);
    return 0;
}

100pts

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;
template<class T> inline void read(T &x) {
    x = 0; RG char c = getchar(); bool f = 0;
    while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
    while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
    x = f ? -x : x;
    return ;
}
template<class T> inline void write(T x) {
    if (!x) {putchar(48);return ;}
    if (x < 0) x = -x, putchar('-');
    int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
    for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}

const int N = 100010;
int a[N], b[N*4], tot, mx[N], mn[N];
int f[N], n, m;


int t[N], p[N];
#define lowbit(x) (x&(-x))
void add(int x, int k) {
    while (x <= tot) {
        t[x] = max(t[x], k);
        x += lowbit(x);
    }
    return ;
}
void clr(int x) {
    while (x <= tot) {
        t[x] = 0;
        x += lowbit(x);
    }
    return ;
}
int query(int x) {
    int s = 0;
    while (x) {
        s = max(s, t[x]);
        x -= lowbit(x);
    }
    return s;
}

inline bool cmp1(int i, int j) { return mx[i] < mx[j]; }
inline bool cmp2(int i, int j) { return a[i] < a[j]; }

void solve(int l, int r) {
    if (l == r) {
        f[l] = max(1, f[l]);
        return ;
    }
    int mid = (l + r) >> 1;
    solve(l, mid);
    for (int i = l; i <= r; i++)
        p[i] = i;
    sort(p+l, p+mid+1, cmp1); sort(p+mid+1, p+r+1, cmp2);
    for (int i = mid+1, j = l; i <= r; i++) {
        while (j <= mid && mx[p[j]] <= a[p[i]]) {
            add(a[p[j]], f[p[j]]);
            j++;
        }
        f[p[i]] = max(f[p[i]], query(mn[p[i]])+1);
    }
    for (int i = l; i <= mid; i++)
        clr(a[i]);
    solve(mid+1, r);
    return ;
}

int main() {
    //freopen(".in", "r", stdin);
    //freopen(".out", "w", stdout);
    read(n); read(m);
    for (int i = 1; i <= n; i++) {
        read(a[i]), mx[i] = mn[i] = a[i];
        b[++tot] = a[i];
    }
    for (int i = 1; i <= m; i++) {
        int x, y; read(x), read(y);
        mx[x] = max(mx[x], y);
        mn[x] = min(mn[x], y);
    }
    for (int i = 1; i <= n; i++) b[++tot] = mx[i], b[++tot] = mn[i];
    sort(b+1, b+1+tot);
    tot = unique(b+1, b+1+tot)-b-1;
    for (int i = 1; i <= n; i++) {
        a[i] = lower_bound(b+1, b+1+tot, a[i])-b;
        mx[i] = lower_bound(b+1, b+1+tot, mx[i])-b;
        mn[i] = lower_bound(b+1, b+1+tot, mn[i])-b;
    }
    solve(1, n);
    int ans = 0;
    for (int i = 1; i <= n; i++)
        ans = max(ans, f[i]);
    printf("%d\n", ans);
    return 0;
}