P1970 花匠

状态定义是dp中非常重要的，可以直接影响到效率，如此题，第一种思路是：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
struct node {
    int high, value;
    bool operator < (const node &i) const{
        return (this->value < i.value);
    }
};
int main() {
    int n;
    scanf("%d", &n);
    int fl[maxn];
    for(int i = 1; i <= n; i++) scanf("%d", &fl[i]);
    int f[maxn], g[maxn];
    priority_queue<node> pf, pg;
    f[1] = 1; g[1] = 1;
    pf.push({fl[1], f[1]}); pg.push({fl[1], g[1]});
    for(int i = 2; i <= n; i++) {
        node x = pf.top();
        node y = pg.top();
        pf.pop(); pg.pop();
        int cnt = 0;
        while(x.high <= fl[i] && cnt < pf.size()) {
            node e = pf.top(); pf.pop();
            pf.push(x); x = e; cnt++;
        }
        cnt = 0;
        while(y.high >= fl[i] && cnt < pg.size()) {
            node e = pg.top(); pg.pop();
            pg.push(y); y = e;cnt++;
        }
        f[i] = 1; g[i] = 1;
        if(x.high > fl[i]) g[i] = x.value + 1;
        if(y.high < fl[i]) f[i] = y.value + 1;
        pf.push({fl[i], f[i]});
        pg.push({fl[i], g[i]});
        pf.push(x);
        pg.push(y);
    }
    int ans = 0;
    for(int i = 1; i <= n; i++) {
        ans = max(ans, f[i]);
        ans = max(ans, g[i]);
    }
    cout << ans;
}

定义f[i]为以i结束，加上优先队列优化，很麻烦，也容易被卡掉，复杂度不稳定，从O(nlogn)~O(n2)

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
int main() {
    int n;
    scanf("%d", &n);
    int fl[maxn];
    for(int i = 1; i <= n; i++) scanf("%d", &fl[i]);
    int f[maxn], g[maxn];
    f[1] = 1; g[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(fl[i] > fl[i-1]) {
            f[i] = max(f[i-1], g[i - 1]+1);
            g[i] = g[i-1];
        }
        else if(fl[i] < fl[i-1]) {
            f[i] = f[i-1];
            g[i] = max(g[i-1], f[i-1] + 1);
        }
        else {
            f[i] = f[i-1];
            g[i] = g[i-1];
        }
    }
    cout << max(f[n], g[n]);
}

定义f[i]为前i个数，不需要优化，效率高，复杂度降低到了O(n)!

算法真神奇。