Balanced Sequence

http://acm.hdu.edu.cn/showproblem.php?pid=6299

Problem Description Chiaki has n strings s1,s2,…,sn consisting of '(' and ')'. A string of this type is said to be balanced: + if it is the empty string + if A and B are balanced, AB is balanced, + if A is balanced, (A) is balanced. Chiaki can reorder the strings and then concatenate them get a new string t . Let f(t) be the length of the longest balanced subsequence (not necessary continuous) of t . Chiaki would like to know the maximum value of f(t) for all possible t .     Input There are multiple test cases. The first line of input contains an integer T , indicating the number of test cases. For each test case: The first line contains an integer n (1≤n≤105 ) -- the number of strings. Each of the next n lines contains a string si (1≤|si|≤105 ) consisting of `(' and `)'. It is guaranteed that the sum of all |si| does not exceeds 5×106 .     Output For each test case, output an integer denoting the answer.     Sample Input   2 1 )()(()( 2 ) )(     Sample Output   4 2

题意：

进行括号匹配，将N个段随意组合，然后进行括号匹配。问最多可以匹配多少括号

分析：

先用栈进行预处理，将n个段最后化简为a个 " ) " 和b个“(”,然后用优先队列优先处理min(a,b)的较大的两组实现贪心，然后将新产生的加入队列。。。

代码：

#include<bits/stdc++.h>
using namespace std;
char s[100005];
struct node
{
    int st,et;
    bool operator<(const node &aa)const
    {
        return min(st,et)<min(aa.st,aa.et);
    }
};
priority_queue<node>q;
stack<char>c;
int main()
{
    int t,i,j,len,sum,aa,bb,tt,n;
    node now,noww;
    scanf("%d",&t);
    while(t--)
    {
        while(!q.empty())q.pop();
        sum=0;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%s",&s);
            len=strlen(s);
            for(j=0;j<len;j++)
            {
                if(s[j]==')')
                {
                    if(!c.empty()&&c.top()=='(')
                        c.pop();
                    else
                        c.push(s[j]);
                }
                else
                c.push(s[j]);
            }
            aa=bb=0;
            while(!c.empty())
            {
                if(c.top()=='(')bb++;
                else
                    aa++;
                c.pop();
            }
            sum+=(len-aa-bb)/2;
            now.st=aa;
            now.et=bb;
            q.push(now);
        }
        while(!q.empty())
        {
            now=q.top();
            q.pop();
            if(q.empty())break;
            noww=q.top();
            q.pop();
            //cout<<"ssss"<<endl;
            //("%d %d %d %d\n",now.st,now.et,noww.st,noww.et);
            if(min(noww.et,now.st)>min(noww.st,now.et))
                swap(now.st,noww.st),swap(now.et,noww.et);
            sum+=min(now.et,noww.st);
            if(now.et>noww.st)
            {
                now.et=now.et-noww.st+noww.et;
            }
            else
            {
                now.st+=noww.st-now.et;
                now.et=noww.et;
            }
            if(now.et||now.st)
            q.push(now);
        }
        printf("%d\n",sum*2);
    }
}