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 + 10;
struct node {
    int l, r, sum;
} qu[maxn];
int cmp(node a, node b) {
    if (a.r <  a.l && b.r >= b.l) return 1;
    if (a.r >= a.l && b.r <  b.l) return 0;
    if (a.r >= a.l && b.r >= b.l)  return a.l > b.l;
    return a.r < b.r;
}
int n, t;
char s[50 * maxn];
int main() {
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        for (int i = 0 ; i < n ; i++) {
            scanf("%s", s);
            qu[i].l = qu[i].r = qu[i].sum = 0;
            int len = strlen(s);
            for (int j = 0 ; j < len ; j++) {
                if (s[j] == '(') qu[i].l++;
                else {
                    if (qu[i].l > 0) qu[i].l--, qu[i].sum++;
                    else qu[i].r++;
                }
            }
        }
        sort(qu, qu + n, cmp);
        int ans = 0, cnt = 0;
        for (int i = 0 ; i < n ; i++) {
            if (qu[i].r > cnt) {
                ans += cnt + qu[i].sum;
                cnt = 0;
            } else {
                ans += qu[i].r + qu[i].sum;
                cnt -= qu[i].r;
            }
            cnt += qu[i].l;
        }
        printf("%d
", ans * 2);
    }
    return 0;
}