D. Least Cost Bracket Sequence
time limit per test
1 second
memory limit per test
64 megabytes
input
standard input
output
standard output
This is yet another problem on regular bracket sequences.
A bracket sequence is called regular, if by inserting "+" and "1" into it we get a correct mathematical expression. For example, sequences "(())()", "()" and "(()(()))" are regular, while ")(", "(()" and "(()))(" are not. You have a pattern of a bracket sequence that consists of characters "(", ")" and "?". You have to replace each character "?" with a bracket so, that you get a regular bracket sequence.
For each character "?" the cost of its replacement with "(" and ")" is given. Among all the possible variants your should choose the cheapest.
Input
The first line contains a non-empty pattern of even length, consisting of characters "(", ")" and "?". Its length doesn't exceed 5·104. Then there follow m lines, where m is the number of characters "?" in the pattern. Each line contains two integer numbers a__i and b__i(1 ≤ a__i, b__i ≤ 106), where a__i is the cost of replacing the i-th character "?" with an opening bracket, and b__i — with a closing one.
Output
Print the cost of the optimal regular bracket sequence in the first line, and the required sequence in the second.
Print -1, if there is no answer. If the answer is not unique, print any of them.
Examples
input
(??)
1 2
2 8
output
4
()()
这题的思路我没想出来,又是看了题解才会的。。
此题思路:http://blog.csdn.net/baidu_29410909/article/details/50967218
ac代码:
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
string s;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
priority_queue
cin>>s;
ll cnt=0,ans=0,m=0,x,y;
for(int i=0;i
ans+=y;
q.push(make_pair(y-x,i));
--cnt;
m++;
}
pair
if(cnt<0) {
if(q.empty()) break;
t=q.top();
q.pop();
ans-=t.first;
s[t.second]='(';
cnt+=2;
}
}
if(cnt!=0) cout<<-1<<endl;
else {
cout<<ans<<endl;
cout<<s<<endl;
}
return 0;
}
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