啊，noip难度题。。。蒟蒻调了半天。。。
嘛、先是求LIS的长度，O(n * logn)算法大家都会
然后就是贪心，假设我们找到了答案的第x项，向后找第x + 1项：
我们发现，只需找当前最前面的a[i]满足f[i] >= l - x的即可。
1 /************************************************************** 2 Problem: 1046 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:1772 ms 7 Memory:960 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 using namespace std; 14 const int N = 10005; 15 const int inf = (int) 1e9; 16 int f[N], a[N], b[N]; 17 int len, n; 18 int ans[N]; 19 20 inline int read(){ 21 int x = 0, sgn = 1; 22 char ch = getchar(); 23 while (ch < '0' || ch > '9'){ 24 if (ch == '-') sgn = -1; 25 ch = getchar(); 26 } 27 while (ch >= '0' && ch <= '9'){ 28 x = x * 10 + ch - '0'; 29 ch = getchar(); 30 } 31 return sgn * x; 32 } 33 34 void pre_work(){ 35 len = 1; 36 int l, r, mid; 37 for (int i = 2; i <= n; ++i) 38 b[i] = -inf; 39 b[0]= inf, b[1] = a[1], f[1] = 1; 40 for (int i = 2; i <= n; ++i){ 41 l = 0, r = len + 1; 42 while (l + 1 < r){ 43 mid = (l + r) >> 1; 44 if (b[mid] > a[i]) l = mid; 45 else r = mid; 46 } 47 len = max(len, ++l); 48 b[l] = max(b[l], a[i]); 49 f[i] = l; 50 } 51 reverse(a + 1, a + n + 1); 52 reverse(f + 1, f + n + 1); 53 } 54 55 void work(int l){ 56 if (len < l){ 57 printf("Impossible "); 58 return; 59 } 60 int cnt = 0, last = -inf; 61 for (int i = 1; i <= n && l != 0; ++i) 62 if (f[i] >= l && a[i] > last) 63 ans[++cnt] = i, --l, last = a[i]; 64 for (int i = 1; i < cnt; ++i) 65 printf("%d ", a[ans[i]]); 66 printf("%d ", a[ans[cnt]]); 67 } 68 69 int main(){ 70 n = read(); 71 for (int i = n; i; --i) 72 a[i] = read(); 73 pre_work(); 74 int m = read(), X; 75 while (m--){ 76 X = read(); 77 work(X); 78 } 79 return 0; 80 }