192B Walking in the Rain

简单DP，设 f[i] 为到达第 i 个位置最迟的天数，那么 f[i] 为 min(f[i-2], a[i]) 和 min(f[i-1], a[i]) 中的较大者；

# include <cstdio>
# include <algorithm>

# define N 1005

using namespace std;

int n, a[N];
int f[N];

void init(void)
{
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
        scanf("%d", &a[i]);
}

void solve(void)
{
    f[1] = a[1], f[2] = min(f[1], a[2]);
    for (int i = 3; i <= n; ++i)
    {
        f[i] = max(min(f[i-2], a[i]), min(f[i-1], a[i]));
    }
    printf("%d\n", f[n]);
}

int main()
{
    init();
    solve();

    return 0;
}