【NOIP2017 DAY1T2】 时间复杂度

【题目链接】

点击打开链接

【算法】

其实这就是一道模拟题啦！

在判error和计算时间复杂度时，我们需要用栈这种数据结构

【代码】

这题的代码还是有些难写的，写的时候一定要有条理！

#include<bits/stdc++.h>
using namespace std;
#define MAXL 100
const int INF = 2e9;

int T,n;
char opt[MAXL+],value[MAXL+];
int l[MAXL+],r[MAXL+];

template <typename T> inline void read(T &x) {
    int f = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
    for (; isdigit(c); c = getchar())  x = (x << ) + (x << ) + c - '';
    x *= f;
}
template <typename T> inline void write(T x) {
    if (x < ) { putchar('-'); x = -x; }
    if (x > ) write(x/);
    putchar(x%+'');
}
template <typename T> inline void writeln(T x) {
    write(x);
    puts("");
}

inline int get() {
    int ret = ;
    bool b = true;
    getchar();
    char c = getchar();
    while (c != ')') {
        if (c == '^') b = false;
        if (isdigit(c)) ret = (ret << ) + (ret << ) + c - '';
        c = getchar();
    }
    getchar();
    if (b) return ;
        else return ret;
}

inline void getstr(int pos) {
        char c = getchar();
        while (c != 'F' && c != 'E') c = getchar();
        opt[pos] = c;
}

inline void getinfo(int pos) {
    int i,len;
    char tx[],ty[];
    getstr(pos);
    if (opt[pos] == 'E') return;
    scanf(" %c %s %s",&value[pos],tx+,ty+);
    if (tx[] != 'n') {
        len = strlen(tx+);
        for (i = ; i <= len; i++) l[pos] = (l[pos] << ) + (l[pos] << ) + tx[i] - '';
    } else
            l[pos] = INF;
    if (ty[] != 'n') {
            len = strlen(ty+);
            for (i = ; i <= len; i++) r[pos] = (r[pos] << ) + (r[pos] << ) + ty[i] - '';
      } else
              r[pos] = INF;
      getchar();
}

inline bool error() {
    int i,top=;
    static int stk[MAXL+];
    static bool used[];
    memset(used,,sizeof(used));
    for (i = ; i <= n; i++) {
        if (opt[i] == 'F') {
            if (used[value[i]-'a']) return true;
            used[value[i]-'a'] = true;
            stk[++top] = i;
        } else {
            if (!top) return true;
            used[value[stk[top]]-'a'] = false;
              --top;
        }
    }
    return top != ;
}

inline void solve() {
    int i,top=,c,tmp,ans=;
    static int stk[MAXL+];
    memset(l,,sizeof(l));
    memset(r,,sizeof(r));
    read(n);
    c = get();
    for (i = ; i <= n; i++) getinfo(i);
    if (error()) {
        puts("ERR");
        return;
    }
    for (i = ; i <= n; i++) {
        if (opt[i] == 'F') {
                tmp = stk[top];
                if (l[i] > r[i]) tmp = -;
                else if (r[i] - l[i] >  && stk[top] != -) ++tmp;
                stk[++top] = tmp;
                ans = max(ans,tmp);
                } else
                        --top;
    }
    if (ans == c) puts("Yes");
    else puts("No");
}

int main() {

    read(T);
    while (T--) solve();

    return ;
}