Yuhao and a Parenthesis

https://codeforces.com/problemset/problem/1097/C

C++版本一

题解：首先保证每个字符串只多(和)其中一个，然后匹配就好了

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=1000000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,q,ans;
int a[N];
char str[N];
stack<char>st;
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endifscanf("%d",&n);memset(a,0,sizeof(a));for(int i=1;i<=n;i++){cin>>str;while (!st.empty()) st.pop();st.push(str[0]);for (int j=1;str[j]!='\0';++j){if (!st.empty()){if (st.top()=='('&&str[j]==')'){st.pop();}else st.push(str[j]);}else st.push(str[j]);}int ls=0,rs=0;while (!st.empty()){char xx=st.top();st.pop();if (xx=='(') ls++;else rs++;}if (ls!=0&&rs!=0){continue;}else if (ls==0&&rs==0){a[500000]++;}else if (rs!=0){a[500000-rs]++;}else a[500000+ls]++;}for(int i=1;i<=500000;i++){ans+=min(a[500000-i],a[500000+i]);}ans+=a[500000]/2;cout << ans << endl;//cout << "Hello world!" << endl;return 0;
}

C++版本二

#include<stdio.h>
#include<string.h>
#include<stack>
using namespace std;
const int mx=5e5+10;
char str[mx];
int num=0;
int sum[mx];
int sum2[mx];
int n;
int ans=0;
stack<char>st;
int main()
{int t;scanf("%d",&t);for (int i=1;i<=t;++i){scanf("%s",str);while (!st.empty()) st.pop();st.push(str[0]);for (int j=1;str[j]!='\0';++j){if (!st.empty()){if (st.top()=='('&&str[j]==')'){st.pop();}else st.push(str[j]);}else st.push(str[j]);}int ls=0,rs=0;while (!st.empty()){char xx=st.top();st.pop();if (xx=='(') ls++;else rs++;}if (ls!=0&&rs!=0){continue;}else if (ls==0&&rs==0){num++;}else if (rs!=0){sum2[rs]++;}else sum[ls]++;}for (int i=1;i<=mx-10;++i){if (sum[i]>sum2[i]){ans+=sum2[i];}else ans+=sum[i];}ans+=num/2;printf("%d\n",ans);
}

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