acwing 187 missile defense system (longest ascending subsequence)

Topic

answer

1. The upgraded version of the interceptor missile, the interceptor missile only allows the use of the non-rising sub-sequence, we only need to consider two cases
   

2. For this question, we can use two sequences of ascending and descending. Then for a number, we have 4 cases, then we can only dfs brute force search to see which case the number is optimal, remember to restore the state

Code

#include<bits/stdc++.h>

using namespace std;
const int N = 55;

int n;
int h[N];
int up[N];  //不下降序列 （序列种元素的值是递减的）
int down[N];  //不上升序列 （序列中元素的值是递增的）
int res;

void dfs(int u, int su, int sd) {
    
    
    //剪枝(已经大于最优解)
    if (su + sd >= res) return;

    if (u == n + 1) {
    
      //已经全部放入序列中
        res = su + sd;
        return;
    }
    //1.将当前数放入上升子序列中
    int k = 0;
    while (k < su && up[k] >= h[u]) k++;
    int t = up[k];
    up[k] = h[u];
    if (k < su) dfs(u + 1, su, sd);
    else dfs(u + 1, su + 1, sd);
    up[k] = t;
    //2.将当前数放入下降子序列中
    k = 0;
    while (k < sd && down[k] <= h[u]) k++;
    t = down[k];
    down[k] = h[u];
    if (k < sd) dfs(u + 1, su, sd);
    else dfs(u + 1, su, sd + 1);
    down[k] = t;
}

int main() {
    
    

    while (cin >> n, n) {
    
    
        for (int i = 1; i <= n; i++) cin >> h[i];
        res = n;  //多组输入,记得更新答案
        dfs(1, 0, 0);
        cout << res << endl;
    }


    return 0;
}