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  • codeforces 3D (非原创)

    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 ai and bi(1 ≤ ai,  bi ≤ 106), where ai is the cost of replacing the i-th character "?" with an opening bracket, and bi — 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 <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <iostream>
    #include <cmath>
    #include <string>
    #include <set>
    #include <queue>
    using namespace std;
    typedef long long ll;
    string s;
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(0);cout.tie(0);
        priority_queue<pair<int,int> >q;
        cin>>s;
        ll cnt=0,ans=0,m=0,x,y;
        for(int i=0;i<s.length();++i) {
            if(s[i]=='(') {
                ++cnt;
            }
            else if(s[i]==')') {
                --cnt;
            }
            else {
                s[i]=')';
                cin>>x>>y;
                ans+=y;
                q.push(make_pair(y-x,i));
                --cnt;
                m++;
            }
            pair<int,int> t;
            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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  • 原文地址:https://www.cnblogs.com/zmin/p/7647651.html
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