G - Two Merged Sequences(dp&贪心)

G - Two Merged Sequences(dp&贪心)

思路1.贪心

维护两个序列，若对于当前元素，只能被加进一个序列中那么直接加入。

若都不能加入输出无解。 否则比较后一个元素。

若 显然加入降序，更优。因为如果加入升序，那么只能加入降序。

而加入降序，后面可以加入更多数，且升序序列也能加入更多数。

若显然加入升序，更优。证明同理。

时间复杂度：

#include <bits/stdc++.h>
using namespace std;

int main() {
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++)
    cin >> a[i];
  int inc = -1, dec = 1e9;
  for (int i = 0; i < n; i++)
    if (inc < a[i] && (a[i] < a[i + 1] || a[i] >= dec))
      inc = a[i], a[i] = 0;
    else if (dec > a[i])
      dec = a[i], a[i] = 1;
    else
      return cout << "NO\n", 0;
  cout << "YES\n";
  for (int i = 0; i < n; i++)
    cout << a[i] << ' ';
}

思路2 dp

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x7f7f7f7f;
int a[200005];
int dp[200005][2];
int op[200005][2];
int opt[200005];
int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    dp[1][1] = inf, dp[1][0] = -inf;
    for (int i = 2; i <= n; i++)
    {
        dp[i][1] = -inf, dp[i][0] = inf;
        if (a[i - 1] < a[i] && dp[i][1] < dp[i - 1][1])
        {
            dp[i][1] = dp[i - 1][1];
            op[i][1] = 1;
        }
        if (dp[i - 1][0] < a[i] && dp[i][1] < a[i - 1])
        {
            dp[i][1] = a[i - 1];
            op[i][1] = 0;
        }
        if (a[i - 1] > a[i] && dp[i][0] > dp[i - 1][0])
        {
            dp[i][0] = dp[i - 1][0];
            op[i][0] = 0;
        }
        if (dp[i - 1][1] > a[i] && dp[i][0] > a[i - 1])
        {
            dp[i][0] = a[i - 1];
            op[i][0] = 1;
        }
    }
    if (dp[n][1] != -inf)
    {
        printf("YES\n");
        for (int i = n, optmp = 1; i >= 1; i--)
        {
            opt[i] = optmp;
            optmp = op[i][optmp];
        }
        for (int i = 1; i <= n; i++)
        {
            printf("%d ", opt[i] != 1);
        }
    }
    else if (dp[n][0] != inf)
    {
        printf("YES\n");
        for (int i = n, optmp = 0; i >= 1; i--)
        {
            opt[i] = optmp;
            optmp = op[i][optmp];
        }
        for (int i = 1; i <= n; i++)
        {
            printf("%d ", opt[i] != 1);
        }
    }
    else
        printf("NO");
    return 0;
}