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 1005using 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;
}

转载于:https://www.cnblogs.com/JMDWQ/archive/2012/07/13/2590898.html

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192B Walking in the Rain

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