[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 = , INF = ~( << );
 int n, ans, max;
 int a[MAXN], f[MAXN][];

 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 = ; i <= n; i++) scanf("%d", &a[i]);
     for(i = ; i <= n; i++)
     {
         max = ;
         for(j = ; j < i; j++)
             if(a[j] < a[i] && f[j][] > max)
                 max = f[j][];
         f[i][] = max + ;
         ans = Max(ans, f[i][]);
         max = ;
         for(j = ; j < i; j++)
             if(a[j] > a[i] && f[j][] > max)
                 max = f[j][];
         f[i][] = max + ;
         ans = Max(ans, f[i][]);
     }
     printf("%d\n", ans);
     return ;
 }

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

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

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

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

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

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

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

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

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

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

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

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 = , INF = ~( << );
 int n, ans;
 int a[MAXN], f[MAXN][];

 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 = ; i <= n; i++) scanf("%d", &a[i]);
     f[][] = f[][] = ;
     for(i = ; i <= n; i++)
     {
         if(a[i] > a[i - ])
         {
             f[i][] = max(f[i - ][], f[i - ][] + );
             f[i][] = f[i - ][];
         }
         else if(a[i] == a[i - ])
         {
             f[i][] = f[i - ][];
             f[i][] = f[i - ][];
         }
         else
         {
             f[i][] = f[i - ][];
             f[i][] = max(f[i - ][], f[i - ][] + );
         }
     }
     printf("%d\n", max(f[n][], f[n][]));
     return ;
 }