时间复杂度
题目背景：
NOIP2017 D1T2
分析：模拟 + 栈
讲真因为T1有点影响心情，考场上打这个题的时候有点混乱，不过好歹最后A了，虽然代码各种不能看，其实就相当于一个人工栈，一个F表示加入，E表示弹出，这样最后可以比较方便的判定非法，然后中途就是判定有多少个n，和多少个常数复杂度。然后判定一下n的个数对不对就好了，因为自己人比较zz，所以实现非常繁琐······
Source
/*created by scarlyw
*/
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cctype>
#include <queue>
#include <vector>template<class T>
inline void R(T &x) {static char c;static bool iosig;for (c = getchar(), iosig = false; !isdigit(c); c = getchar()) {if (c == -1) return ;if (c == '-') iosig = true;}for (x = 0; isdigit(c); c = getchar())x = ((x << 2) + x << 1) + (c ^ '0');if (iosig) x = -x;
}char c[10], temp;
bool flag;
bool vis[30];
int x, y, cnt, l, t;inline int dfs(int cnt, int num, int cc) {int max = 0;char temp = 0;while (true) {temp = 0;while (temp != 'F' && temp != 'E') temp = getchar();::cnt++;if (temp == 'E') {vis[cc] = false;return max * num + cnt;}char x = 0;while (x > 'z' || x < 'a') x = getchar();if (vis[x - 'a'] == true) flag = false;else vis[x - 'a'] = true;char a[10], b[10];scanf("%s%s", a, b);int d = -1;if (a[0] == 'n' && b[0] == 'n') d = 1;else if (a[0] == 'n' && b[0] != 'n') d = -1;else if (a[0] != 'n' && b[0] == 'n') d = 2;else if (a[0] != 'n' && b[0] != 'n') {int s = 0, t = 0, len;len = strlen(a);for (int j = 0; j < len; ++j) s = (s * 10) + (a[j] ^ '0');len = strlen(b);for (int j = 0; j < len; ++j) t = (t * 10) + (b[j] ^ '0');if (s <= t) d = 1;else d = -1;}if (::cnt == l) {flag = false;return 0;}if (d == -1) max = std::max(max, dfs(0, 0, x - 'a'));else if (d == 1) max = std::max(max, dfs(0, 1, x - 'a'));else if (d == 2) max = std::max(max, dfs(1, 1, x - 'a'));if (::cnt == l) {flag = false;return 0;}}
}inline void solve() {R(l), flag = true, cnt = 0, temp = 0;memset(vis, 0, sizeof(vis));while (temp != '1' && temp != 'n') temp = getchar();if (temp == '1') y = 0;else R(y);int ans = 0;for (cnt = 1; cnt <= l; cnt++) {temp = 0;while (temp != 'F' && temp != 'E') temp = getchar();if (temp == 'E') {flag = false;continue ;}char x = 0;while (x > 'z' || x < 'a') x = getchar();if (vis[x - 'a'] == true) flag = false;else vis[x - 'a'] = true;char a[10], b[10];scanf("%s%s", a, b);int d = -1;if (a[0] == 'n' && b[0] == 'n') d = 1;else if (a[0] == 'n' && b[0] != 'n') d = -1;else if (a[0] != 'n' && b[0] == 'n') d = 2;else if (a[0] != 'n' && b[0] != 'n') {int s = 0, t = 0, len;len = strlen(a);for (int j = 0; j < len; ++j) s = (s * 10) + (a[j] ^ '0');len = strlen(b);for (int j = 0; j < len; ++j) t = (t * 10) + (b[j] ^ '0');if (s <= t) d = 1;else d = -1;}if (cnt == l) {flag = false;break ;}if (d == -1) ans = std::max(ans, dfs(0, 0, x - 'a'));else if (d == 1) ans = std::max(ans, dfs(0, 1, x - 'a'));else ans = std::max(ans, dfs(1, 1, x - 'a'));if (cnt == l) break ;}if (flag == false) std::cout << "ERR\n";else if (ans == y) std::cout << "Yes\n";else std::cout << "No\n";
}int main() {freopen("complexity.in", "r", stdin);freopen("complexity.out", "w", stdout);R(t);while (t--) solve();return 0;
}