洛谷P3952 時間復雜度(模擬)
阿新 • • 發佈:2018-10-31
str turn har 鏈接 strlen 是個 etop round 自己
題意
題目鏈接
Sol
咕了一年的題解。。就是個模擬吧
考場上寫的遞歸也是醉了。。。
感覺一年自己進步了不少啊。。面向數據編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 是否算作復雜度 1 算 0不算 void init() { top = 0;//已經用過哪些字符 now = 0;//當前進入了幾層循環 mx = 0;//最大循環層數 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常數 -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;}//不進入循環 if(bg != -1 && ed == -1) {//非常數循環 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 */
洛谷P3952 時間復雜度(模擬)