hdu 6299 Balanced Sequence （贪心）

Balanced Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6311    Accepted Submission(s): 1648

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

 

题目大意：

给你n个由'('和')'组成的括号序列。要你将这n个序列整体排序使得匹配的子序列（不一定相连）最长。求最长子序列。匹配条件如下：

+ if it is the empty string
+ if A and B are balanced, AB is balanced,
+ if A is balanced, (A) is balanced.

 

贪心。

首先将每个序列中匹配的括号计数并剔除。可以看到，剩下的序列都行如")))((((("，用结构体存储r，l，排序，就可以贪心了。

贪心方法是（其实我并不能证明，大概是基于左括号尽量往左放，vice versa的思想）：

1、"))))((((" 中 ')' < '(' 的 , 按 ')' 从小到大排序 ;

2、"))))((((" 中 ')' >= '(' 的 , 按 '(' 从大到小排序 ;

 

//优先级排序：
//1、"))))((((" 中 ')' < '(' 的 , 按 ')' 从小到大排序 ;
//2、"))))((((" 中 ')' >= '(' 的 , 按 '(' 从大到小排序 ;

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<stack>

using namespace std;

const int maxn=100000;

char s[maxn+10];

struct tstr
{
    int r,l;
};
tstr str[maxn+10];

bool cmp(tstr a,tstr b)
{
    if(a.r<a.l&&b.r>=b.l)
        return true;
    if(b.r<b.l&&a.r>=a.l)
        return false;
    if(a.r<a.l&&b.r<b.l)
        return a.r<b.r;
    else
        return a.l>b.l;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);

        int ans=0;
        for(int i=0;i<n;i++)
        {
            scanf("%s",s);
            stack<char> sta;
            sta.push(')');//在栈底放一个')'，方便后续操作
            for(int j=0;s[j]!='';j++)
            {
                if(s[j]==')'&&sta.top()=='(')
                    ans+=2,sta.pop();
                else
                    sta.push(s[j]);
            }
            str[i].r=-1;str[i].l=0;
            while(!sta.empty())
            {
                if(sta.top()=='(')
                    str[i].l++,sta.pop();
                if(sta.top()==')')
                    str[i].r++,sta.pop();
            }
        }

        sort(str,str+n,cmp);
        stack<char> sta;
        sta.push(')');
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<str[i].r;j++)
            {
                if(sta.top()=='(')
                    ans+=2,sta.pop();
                else
                    sta.push(')');
            }
            for(int j=0;j<str[i].l;j++)
            {
                sta.push('(');
            }
        }

        printf("%d
",ans);
    }
    return 0;
}