[luoguP1970] 花匠（DP）

传送门

n² 过不了惨啊

70分做法

f[i][0] 表示第 i 个作为高的，的最优解

f[i][0] 表示第 i 个作为低的，的最优解

（且第 i 个一定选）

那么

f[i+1][1]=max(f[j][0])+1,i>=j>=1,h[j]>h[i+1],

f[i+1][0]=max(f[j][1])+1,i>=j>=1,h[j]<h[i+1],

——代码

#include <cstdio>
#include <algorithm>

const int MAXN = 100001, INF = ~(1 << 31);
int n, ans, max;
int a[MAXN], f[MAXN][2];

inline int Max(int x, int y)
{
    return x > y ? x : y;
}

int main()
{
    //freopen("flower.in", "r", stdin);
    //freopen("flower.out", "w", stdout);
    int i, j;
    scanf("%d", &n);
    for(i = 1; i <= n; i++) scanf("%d", &a[i]);
    for(i = 1; i <= n; i++)
    {
        max = 0;
        for(j = 1; j < i; j++)
            if(a[j] < a[i] && f[j][1] > max)
                max = f[j][1];
        f[i][0] = max + 1;
        ans = Max(ans, f[i][0]);
        max = 0;
        for(j = 1; j < i; j++)
            if(a[j] > a[i] && f[j][0] > max)
                max = f[j][0];
        f[i][1] = max + 1;
        ans = Max(ans, f[i][1]);
    }
    printf("%d
", ans);
    return 0;
}

100分

我们发现上面的方程可以用线段树优化，可以建一颗权值线段树

#include <cstdio>
#include <iostream>
#include <algorithm>
#define root 1, 1, size
#define ls now << 1, l, mid
#define rs now << 1 | 1, mid + 1, r

const int MAXN = 100001;
int n, size;
int a[MAXN], b[MAXN], f[MAXN][2], max[MAXN << 2][2];

inline int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
    for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
    return x * f;
}

inline int Max(int x, int y)
{
    return x > y ? x : y;
}

inline void pushup(int now)
{
    max[now][0] = Max(max[now][0], max[now << 1][0]);
    max[now][0] = Max(max[now][0], max[now << 1 | 1][0]);
    max[now][1] = Max(max[now][1], max[now << 1][1]);
    max[now][1] = Max(max[now][1], max[now << 1 | 1][1]);
}

inline int query(int x, int y, int k, int now, int l, int r)
{
    if(x <= l && r <= y) return max[now][k];
    int mid = (l + r) >> 1;
    if(l > y || r < x) return 0;
    return Max(query(x, y, k, ls), query(x, y, k, rs));
}

inline void update(int x, int a, int b, int now, int l, int r)
{
    if(l == r)
    {
        max[now][0] = Max(max[now][0], a);
        max[now][1] = Max(max[now][1], b);
        return; 
    }
    int mid = (l + r) >> 1;
    if(x <= mid) update(x, a, b, ls);
    else update(x, a, b, rs);
    pushup(now);
}

int main()
{
    int i;
    n = read();
    for(i = 1; i <= n; i++) a[i] = b[i] = read();
    std::sort(b + 1, b + n + 1);
    size = std::unique(b + 1, b + n + 1) - (b + 1);
    for(i = 1; i <= n; i++) a[i] = std::lower_bound(b + 1, b + size + 1, a[i]) - b;
    for(i = 1; i <= n; i++)
    {
        f[i][0] = query(1, a[i] - 1, 1, root) + 1;
        f[i][1] = query(a[i] + 1, size, 0, root) + 1;
        update(a[i], f[i][0], f[i][1], root);
    }
    printf("%d
", Max(f[n][0], f[n][1]));
    return 0;
}

100分

用个p线段树，树状数组维护前后缀就好。

#include <cstdio>
#include <iostream>
#include <algorithm>
#define root 1, 1, size
#define ls now << 1, l, mid
#define rs now << 1 | 1, mid + 1, r

const int MAXN = 1000002;
int n, ans;
int c0[MAXN], c1[MAXN];

inline int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
    for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
    return x * f;
}

inline int max(int x, int y)
{
    return x > y ? x : y;
}

inline int query1(int x)
{
    int ret = 0;
    for(; x <= MAXN; x += x & -x) ret = max(ret, c1[x]);
    return ret;
}

inline int query0(int x)
{
    int ret = 0;
    for(; x; x -= x & -x) ret = max(ret, c0[x]);
    return ret;
}

inline void update0(int x, int d)
{
    for(; x <= MAXN; x += x & -x) c0[x] = max(c0[x], d);
}

inline void update1(int x, int d)
{
    for(; x; x -= x & -x) c1[x] = max(c1[x], d);
}

int main()
{
    int i, x, y, z;
    n = read();
    for(i = 1; i <= n; i++)
    {
        z = read() + 1;
        x = query1(z + 1) + 1;
        y = query0(z - 1) + 1;
        update0(z, x);
        update1(z, y);
        ans = max(ans, max(x, y));
    }
    printf("%d
", ans);
    return 0;
}

100分

f[i][0] 表示第 i 个作为高的，的最优解

f[i][0] 表示第 i 个作为低的，的最优解

（然而第 i 个不一定选）

那么

h[i]>h[i−1]时， 

f[i][0]=max{f[i−1][0],f[i−1][1]+1},f[i][1]=f[i−1][1]; 

h[i]==h[i−1]时， 

f[i][0]=f[i−1][0],f[i][1]=f[i−1][1]; 

h[i]<h[i−1]时， 

f[i][0]=f[i−1][0],f[i][1]=max{f[i−1][1],f[i−1][0]+1}. 

答案ans=max{f[n][0],f[n][1]}; 

边界为f[1][0]=f[1][1]=1

——代码

#include <cstdio>
#include <algorithm>

const int MAXN = 100001, INF = ~(1 << 31);
int n, ans;
int a[MAXN], f[MAXN][2];

inline int max(int x, int y)
{
    return x > y ? x : y;
}

int main()
{
    //freopen("flower.in", "r", stdin);
    //freopen("flower.out", "w", stdout);
    int i, j;
    scanf("%d", &n);
    for(i = 1; i <= n; i++) scanf("%d", &a[i]);
    f[1][0] = f[1][1] = 1;
    for(i = 2; i <= n; i++)
    {
        if(a[i] > a[i - 1])
        {
            f[i][0] = max(f[i - 1][0], f[i - 1][1] + 1);
            f[i][1] = f[i - 1][1];
        }
        else if(a[i] == a[i - 1])
        {
            f[i][0] = f[i - 1][0];
            f[i][1] = f[i - 1][1];
        }
        else
        {
            f[i][0] = f[i - 1][0];
            f[i][1] = max(f[i - 1][1], f[i - 1][0] + 1);
        }
    }
    printf("%d
", max(f[n][0], f[n][1]));
    return 0;
}