Balanced Sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5225 Accepted Submission(s): 1339
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 Source 2018 Multi-University Training Contest 1 Recommend liuyiding | We have carefully selected several similar problems for you: 6331 6330 6329 6328 6327 #include<bits/stdc++.h>
#define maxn 1000010
using namespace std;
int ans,len,cnt;
int ll,a,b,m; struct node
{
int l,r;///左括号和右括号的数量
};
node s[maxn];
/*
题目大意：给定数个括号字符串，问如何拼接使得出现的()最多。 首先把原来的字符串预先处理完后，发现只剩下三种情况，
全')',全'(',和两种拼接而成的。
在排序时可以考虑，把其中数量多的当成单个括号，
比如如果'('比')'多，不妨把整个序列考虑优先级时考虑成'('的优先级。
当一样的时候手动模拟一下找找规律即可。
*/
bool cmp(node a,node b)
{
if(a.l<=a.r&&b.l>b.r) return true;
if(a.l>a.r&&b.l<=b.r) return false;
if(a.l>=a.r && b.l>b.r) return a.r>b.r;
if(a.l<=b.r&&b.l<=b.r) return a.l<b.l;
} int main()
{
int t;scanf("%d",&t); char seq[maxn],sk[maxn];
int top; while(t--)
{
scanf("%d",&m); ans=0;
top=0;
for(int i=0;i<m;i++)
{
scanf("%s",seq);
/*
cnt=0 , len=strlen(seq);///cnt是匹配括号的数量
ll=0 , a=0 , b=0 ;///ll用于配合计算cnt
*/
len=strlen(seq);
for(int j=0;j<len;j++)
{
if(seq[j]==')'&&sk[top]=='('&&top>0)
{
ans++;
top--;
}
else sk[++top]=seq[j];
}
s[i].l=s[i].r=0;
for(int j=1;j<=top;j++)
{
if(sk[j]=='(') s[i].r++;
else s[i].l++;
}
}
sort(s,s+m,cmp);
int num=0;
for(int i=1;i<m;i++)
{
num+=s[i-1].r;
if(s[i].l>=num)
{
ans+=num;
num=0;
}
else
{
num-=s[i].l;
ans+=s[i].l;
}
}
printf("%d\n",ans*2);
}
return 0;
}
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