Yuhao and a Parenthesis
阿新 • • 發佈:2019-01-07
https://codeforces.com/problemset/problem/1097/C
C++版本一
題解:首先保證每個字串只多(和)其中一個,然後匹配就好了
/* *@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 DEBUG freopen("input.in", "r", stdin); //freopen("output.out", "w", stdout); #endif scanf("%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); }