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

C++版本一

題解：首先保證每個字符串只多(和)其中一個，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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