如果要看能统计方案数的加强版,可以转至划分。
这题如果每次暴力从上一个状态枚举转移可以得到一个时间复杂度为 \(O(n^2)\) 的朴素动态规划,期望得分 \(19\) 分。
观察朴素动态规划方程,并经过打表可以发现,对于每一个阶段,在一个序列中选取点的个数是连续的一段区间。
那么我们可以考虑少枚举一层状态(选取个数),把控制区间记录在数组里,这样就可以 \(O(1)\) 转移啦。
输出答案方案,只需要逆序判断就可。
至于我上文提到的加强版,需要读者掌握一定的多项式技术,笔者也在自己研究中,可以参考牛客网出题人的题解。
code:
// Program written by Liu Zhaozhou ~~~
#include <bits/stdc++.h>using namespace std;inline char gc(void) {static char buf[100000], *p1 = buf, *p2 = buf;return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;
}template <class T> inline void read(T &x) {T f = 1; x = 0; static char c = gc();for (; !isdigit(c); c = gc()) if (c == '-') f = -f;for (; isdigit(c); c = gc()) x = (x << 1) + (x << 3) + (c & 15);x *= f;
}inline void readstr(string&x) {x = ""; static char ch;while (isspace(ch = gc()));while (x += ch, !isspace(ch = gc()));
}inline void readstr(char *s) {do *s = gc(); while ((*s == ' ') || (*s == '\n') || (*s == '\r'));do *(++s) = gc(); while ((~*s) && (*s != ' ') && (*s != '\n') && (*s != '\r'));*s = 0; return;
}inline void readch(char&x) { while (isspace(x = gc())); }char pf[100000], *o1 = pf, *o2 = pf + 100000;
#define ot(x) (o1 == o2 ? fwrite(pf, 1, 100000, stdout), *(o1 = pf) ++= x : *o1 ++= x)
template <class T>
inline void println(T x, char c = '\n') {if (x < 0) ot(45), x = -x;static char s[15], *b; b = s;if (!x) *b ++= 48;for (; x; * b ++= x % 10 + 48, x /= 10);for (; b-- != s; ot(*b)); ot(c);
}template <class T> inline void write(T x) {if (x < 0) x = -x, putchar('-');if (x > 9) write(x / 10);putchar(x % 10 + 48);
}template <class T> inline void writeln(T x, char c = '\n') { write(x); putchar(c); }
template <class T> inline void chkmax(T &x, const T y) { x > y ? x = x : x = y; }
template <class T> inline void chkmin(T &x, const T y) { x < y ? x = x : x = y; }inline void file(string str) {freopen((str + ".in").c_str(), "r", stdin);freopen((str + ".out").c_str(), "w", stdout);
}#define Ms(arr, opt) memset(arr, opt, sizeof(arr))
#define Mp(x, y) make_pair(x, y)typedef long long ll;
typedef pair <int, int> pii;const int INF = 0x3f3f3f3f;const int Maxn = 5e5 + 5;
int n, a[Maxn * 2], b[Maxn * 2];
pii f[Maxn * 2][2]; char ans[Maxn * 2];
inline void update(pii &x, pii y) {chkmin(x.first, y.first);chkmax(x.second, y.second);
}signed main(void) {read(n); int m = n * 2;for (int i = 1; i <= m; i++) read(a[i]);for (int i = 1; i <= m; i++) read(b[i]);f[0][0] = f[0][1] = Mp(0, 0);for (int i = 1; i <= m; i++) {f[i][0] = f[i][1] = Mp(INF, -INF);if (a[i - 1] <= a[i]) update(f[i][0], f[i - 1][0]);if (a[i - 1] <= b[i]) update(f[i][1], f[i - 1][0]);if (b[i - 1] <= a[i]) update(f[i][0], f[i - 1][1]);if (b[i - 1] <= b[i]) update(f[i][1], f[i - 1][1]);++f[i][1].first; ++f[i][1].second;}int cur = -1, rem = n;for (int i = 0; i < 2; i++)if (f[m][i].first <= n && f[m][i].second >= n) cur = i;if (cur == -1) { puts("-1"); return 0; }for (int i = m; i >= 1; i--) {ans[i] = (cur ? 'B' : 'A'); rem -= cur;int opt = (cur ? b[i] : a[i]);if (a[i - 1] <= opt && f[i - 1][0].first <= rem && f[i - 1][0].second >= rem) cur = 0;else cur = 1;} puts(ans + 1);
// fwrite(pf, 1, o1 - pf, stdout);return 0;
}/**/