Balanced Sequence（毒瘤啊）排序贪心 HDU多校

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

 

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int maxn = 1e5 + ;
 struct node {
     int l, r, sum;
 } qu[maxn];
 int cmp(node a, node b) {
     if (a.r <  a.l && b.r >= b.l) return ;
     if (a.r >= a.l && b.r <  b.l) return ;
     if (a.r >= a.l && b.r >= b.l)  return a.l > b.l;
     return a.r < b.r;
 }
 int n, t;
 char s[ * maxn];
 int main() {
     scanf("%d", &t);
     while(t--) {
         scanf("%d", &n);
         for (int i =  ; i < n ; i++) {
             scanf("%s", s);
             qu[i].l = qu[i].r = qu[i].sum = ;
             int len = strlen(s);
             for (int j =  ; j < len ; j++) {
                 if (s[j] == '(') qu[i].l++;
                 else {
                     if (qu[i].l > ) qu[i].l--, qu[i].sum++;
                     else qu[i].r++;
                 }
             }
         }
         sort(qu, qu + n, cmp);
         int ans = , cnt = ;
         for (int i =  ; i < n ; i++) {
             if (qu[i].r > cnt) {
                 ans += cnt + qu[i].sum;
                 cnt = ;
             } else {
                 ans += qu[i].r + qu[i].sum;
                 cnt -= qu[i].r;
             }
             cnt += qu[i].l;
         }
         printf("%d\n", ans * );
     }
     return ;
 }