#define inf 0x3f3f3f3f
#define ll long long
const int maxn = 1e4 + 500;
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
//最长上升子序列问题)(nlogn)
int dp[maxn];//dp[i]表示最优子序列到第i位置的最小值(为了后面最优长度做铺垫)
int a[maxn];
int pos[maxn];//记录a[i]存放/交换dp[]中的位置(求最优字串才需要)
int n;
void solve()
{
fill(dp, dp + n, inf);
for (int i = 0; i < n; i++)
{
pos[i] = lower_bound(dp, dp + n, a[i]) - dp; //记录位置
*lower_bound(dp, dp + n, a[i]) = a[i]; //在dp[0]~dp[n-1]中二分找到第一个大于等于a[i]的位置并交换为a[i]
}
int maxlen = lower_bound(dp, dp + n, inf) - dp;
int t = maxlen - 1;//从0开始
for (int i = n - 1; i >= 0; i--)
{
if (pos[i] == t)dp[t--] = a[i];
if (t < 0)break;
}
printf("%d\n", maxlen);
for (int i = 0; i < maxlen; i++)
{
printf("%d", dp[i]);
if (i != maxlen - 1) printf(" ");
}
printf("\n");
}
int main()
{
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%d", &a[i]);
solve();
return 0;
}
DP之最长上升子序列模板O(nlogn) (最长上升子序列)
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转载自blog.csdn.net/weixin_41156591/article/details/81810477
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