CodeForces - 3D（优先队列）

Poj CodeForces - 3D

?. Least Cost Bracket Sequence:

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
()()
思路：先把问号‘？’看做右括号，然后判断左括号个数是否小于右括号，如果小，就从优先队列里拿出权值最小的左括号。。。
AC代码：

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
typedef long long ll;
const int N = 1e5+10;
struct node{
    int num, id;
    bool operator < (const node & a) const{
        return num > a.num;
    }
};
char s[N];
int main(){
    #ifdef ONLINE_JUDGE
    #else
        freopen("in.txt", "r", stdin);
    #endif // ONLINE_JUDGE
    int x, y;
    while(~scanf("%s", s)){
        int len = strlen(s);
        int l = 0, r = 0;
        ll ans = 0;
        bool flag = false;
        priority_queue<node> Q;
        for(int i = 0; i < len; i++){
            if(s[i] == '(') l++;
            else if(s[i] == ')') r++;
            else{
                scanf("%d%d", &x, &y);
                ans += y;
                Q.push((node){x - y, i});
                r++;
            }
            if(l < r){
                if(Q.empty()) flag = true;
                else {
                    ans += Q.top().num;
                    s[Q.top().id] = '(';
                    Q.pop();
                    l++;
                    r--;
                }
            }
        }
        if(flag || l != r) printf("-1");
        else{
            printf("%lld\n", ans);
            for(int i = 0; i < len; i++){
                if(s[i] == '?') printf(")");
                else printf("%c", s[i]);
            }
        }
        printf("\n");
    }
    return 0;
}