洛谷P3952 时间复杂度(模拟)

题意

题目链接

Sol

咕了一年的题解。。就是个模拟吧

考场上写的递归也是醉了。。。

感觉一年自己进步了不少啊。。面向数据编程的能力提高了不少

#include<bits/stdc++.h>
#define fi first
#define se second
#define MP make_pair
using namespace std;
const int MAXN = 101;
int T, top = 0, now, mx, flag;
pair<char, int> st[MAXN];// first 字符  second 是否算作复杂度 1 算 0不算
void init() {
	top = 0;//已经用过哪些字符
	now = 0;//当前进入了几层循环
	mx = 0;//最大循环层数
	flag = 0;
}
int get(char *s) {
	int l = strlen(s + 1), x = 0;
	for(int i = 1; i <= l; i++) if(s[i] >= '0' && s[i] <= '9') x = x * 10  + s[i] - '0';
	return x;
}
char getopt() {
	char c = ' ';
	while(c != 'E' && c != 'F') c = getchar();
	return c == 'F' ? 1 : 0;// 1 enter  0 end
}
int readround() {//n = -1
	char c = ' '; int x = 0;
	while(c != 'n' && (c < '0' || c > '9')) c = getchar();
	if(c == 'n') return -1;
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x;
}
int readbuf() {//  1 n  0常数   -1重名
	char c = getchar();
	while(c < 'a' || c > 'z') c = getchar();
	int bg = readround(), ed = readround();
	for(int i = 1; i <= top; i++) if(st[i].fi == c) return -1;
	if((bg != -1 && ed != -1 && bg > ed) || (bg == -1 && ed != -1)) {flag = 1, st[++top] = MP(' ', -1); return 0;}//不进入循环
	if(bg != -1 && ed == -1) {//非常数循环
		if(flag == 0) now++, mx = max(now, mx), st[++top] = MP(c, 1);
		else st[++top] = MP(c, 0);
	} else {
		st[++top] = MP(c, 0);
	}
	return 0;
}
int solve() {// 1 Yes 0 No  -1 ERR
	int L, w = 0, GG = 0; char s[233];
	scanf("%d %s", &L, s + 1);
	if(s[3] == 'n') w = get(s + 1);
	for(int i = 1; i <= L; i++) {
		int opt = getopt();
		if(opt == 0) {
			if(top == 0) GG = -1;
			else {
				if(st[top].se == -1) flag = 0;
				if(st[top--].se == 1) now--;
			}
		} else {
			int tmp = readbuf();
			if(tmp == -1) GG = -1;
		}
	}
	if(GG == -1) return -1;
	if(top) return -1;
	else return mx == w;
}
int main() {
	cin >> T;
	while(T--) {
		init();
		int tmp = solve();
		if(tmp == 1) puts("Yes");
		else if(tmp == 0) puts("No");
		else puts("ERR");
	}
	return 0;
}
/*
1
2 O(n^1)
F a n n
E

*/