有点贪心

 

You are given nn strings str1,str2,…,strnstr1,str2,…,strn, each consisting of ( and ). The objective is to determine whether it is possible to permute the nn strings so that the concatenation of the strings represents a valid string.

Validity of strings are defined as follows:

• The empty string is valid.
• If AA and BB are valid, then the concatenation of AA and BB is valid.
• If AA is valid, then the string obtained by putting AA in a pair of matching parentheses is valid.
• Any other string is not valid.

For example, "()()" and "(())" are valid, while "())" and "((()" are not valid.

Input

The first line of the input contains an integer nn (1≤n≤1001≤n≤100), representing the number of strings. Then nn lines follow, each of which contains stristri (1≤|stri|≤1001≤|stri|≤100). All characters in stristri are ( or ).

Output

Output a line with "Yes" (without quotes) if you can make a valid string, or "No" otherwise.

Sample Input 1

3
()(()((
))()()(()
)())(())

Output for the Sample Input 1

Yes

Sample Input 2

2
))()((
))((())(

Output for the Sample Input 2

No

//  思维+贪心
// Aizu - 2681 
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <cstdlib>
typedef long long ll;
#define lowbit(x) (x&(-x))
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
using namespace std;
const int N=1e2+9;
int n;
char s[N];
typedef pair<int,int>P;
P p[N];
P solve(char s[])
{
    int l=0,r=0;
    int i;
    for(i=0;s[i];i++){
        if(s[i]=='(')  l++;
        else if(l)  l--;
        else r++;
    }
    return P(l,r);//l为( 数目，r为)数目
}
bool vis[N];
int main()
{
    scanf("%d",&n);
    int  cnt=0;
    for(int i=0;i<n;i++)
    {
        scanf("%s",s);
        p[i]=solve(s);
        if(!p[i].second) cnt+=p[i].first,i--,n--;
        // cnt 为只有(的字符串的（的和
    }
     int ans=0,ret=1;
     memset(vis,0,sizeof(vis));
     int mx,id;
     for(int i=0;i<n&&ret;i++){
         id=-1,mx=-120;
         for(int j=0;j<n;j++)
         {
             if(vis[j]||ans<p[j].first) continue;
             int k=p[j].second-p[j].first;
             if(mx<k) mx=k,id=j;
         }
         //第一个放在最右边 ，依次往左（前）放。
        // mx 提供最多的右括号，（贪心）
         if(id==-1)  ret=0;
         vis[id]=1;
         ans+=mx;//ans 为这个字符串的净提供右括号数目。
     }
    // printf("%d %d %d\n",ret,ans,cnt);
     if(ret&&ans==cnt)  printf("Yes\n");
     else printf("No\n");
     return 0;
}