2016 Multi-University Training Contest 8 1011 Rikka with Parenthesis II(模拟)
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题意
给出一堆括号,要求必须交换其中的两个,问交换后是否合法。
思路
很显然,当恰好有一对错位时,交换合法,当原来本身合法 且数量大于4时。可以合法交换,如()是不合法的。扫描的同时记录待匹配的(的个数为cnt,遇到)则减1,若min(cnt) <= -2 则无解。输入为奇数也无解,最后()的数量不匹配也无解。细心模拟就好了
代码
#include <bits/stdc++.h>using namespace std;const int maxn = 1e5+10;char str[maxn];int main (){ int T; scanf("%d",&T); while(T --){ int n; scanf("%d", &n); scanf("%s",str); int cnt = 0; int cc = 0; for(int i = 0 ; i < n ; i ++){ if(str[i] == '(') cnt ++; else cnt--; if(cnt < 0) cc = max(-1*cnt,cc); } bool flag = false; if (cc <= 2 && cnt == 0) flag = true; if (str[0] == '(' && cnt == 0 && n == 2) flag = false; if(flag) puts("Yes"); else puts("No"); }}
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