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

*/