CodeForces 3D. Least Cost Bracket Sequence

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.

Sample test(s)
input
(??)
1 2
2 8
output
4
()()

贪心法：先不考虑括号匹配，将所有问号设置为值较小的括号。

再检查总的括号数是多了左括号还是多了右括号。

如果多了右括号：设需要更换为左括号的右括号数量为numbers，此时data从左向右循环，如果遇到左括号sum++，遇到右括号sum--，如果sum<0就在 I 左侧寻找最优的右括号 更换为左括号，sum+=2。

如果多了左括号，原理同上，只需data从右向左，如果sum>0,寻找最优左括号，sum-=2。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char data[50004];
int a[50004],b[50004],num[50004],len,cur;
long long sum;
int main(){
    endw:
    while(scanf("%s",data)!=EOF){
        sum=cur=0;
        int SumUnsure=0;
        len=strlen(data);

        for(int i=0;i<len;i++){
            if(data[i]=='(') { sum++;data[i]=1; }
            else if(data[i]==')') { sum--; data[i]=-1;}
            else if(data[i]=='?') {
                scanf("%d%d",&a[i],&b[i]);
                if(a[i]<=b[i]) {data[i]=1;SumUnsure++;}//<span style="color: rgb(34, 34, 34); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 19.6000003814697px;">将所有问号设置为值较小的括号。</span>
                else if(b[i]<a[i]) {data[i]=-1;SumUnsure--;}
                num[cur++]=i;
            }
        }
        if(len%2==1) {printf("-1\n");continue;}
        int type=0,numbers=0;
        if(SumUnsure+sum!=0){
            if(SumUnsure+sum<0) {type=-1;numbers=-(SumUnsure+sum)/2;}
            else {type=1;numbers=(SumUnsure+sum)/2;}
        }
        sum=0;
        if(type==-1||type==0){
        for(int I=0;I<len;I++){//从第一个元素往后检测，更换多余的右括号
            sum+=data[I];
            if(sum<0) {
                int AminR_L=1000000,AminR_L_num;
                for(int j=0,i=num[0];i<=I&&j<cur;i=num[++j]){ //寻找 差值最小 的右括号
                    if(data[i]==-1&&AminR_L>a[i]-b[i]) {
                       AminR_L = a[i]-b[i];
                       AminR_L_num = i;
                    }
                }
                if(AminR_L==1000000) { printf("-1\n");goto endw; }
                int AminL_R=1000000,AminL_R_num;
                if(numbers==0){
                    for(int j=cur-1,i=num[cur-1];i>I;i=num[--j]) { //寻找 差值最小 的左括号
                        if(data[i]==1&&AminL_R>b[i]-a[i]) {
                            AminL_R = b[i]-a[i];
                            AminL_R_num = i;
                        }
                        if(j<=0) break;
                    }
                    if(AminL_R==1000000) { printf("-1\n");goto endw; }
                }
                else numbers--;

                if(AminL_R==1000000) {//numbers!=0
                    data[AminR_L_num] = -data[AminR_L_num];

                }
                else {
                    data[AminR_L_num] = -data[AminR_L_num];
                    data[AminL_R_num] = -data[AminL_R_num];
                }
                sum+=2;
            }
        }
        }
        else if(type==1){
        for(int I=len-1;I>=0;I--){//从最后一个元素往前检测，更换多余的左括号
            sum+=data[I];
            if(sum>0) {
                int Amin1=1000000,Amin1_num;

                for(int j=cur-1,i=num[cur-1];i>=I;i=num[--j]) { //寻找 差值最小 的左括号
                    if(data[i]==1&&Amin1>b[i]-a[i]) {
                        Amin1 = b[i]-a[i];
                        Amin1_num = i;
                    }
                    if(j<=0) break;
                }

                if(Amin1==1000000) { printf("-1\n");goto endw; }

                int Amin2=1000000,Amin2_num;
                if(numbers==0){
                    for(int j=0,i=num[0];i<I;i=num[++j]){ //寻找 差值最小 的右括号
                        if(data[i]==-1&&Amin2>a[i]-b[i]) {
                            Amin2 = a[i]-b[i];
                            Amin2_num = i;
                        }
                        if(j>=cur-1) break;
                    }
                    if(Amin2==1000000) { printf("-1\n");goto endw; }
                }
                else numbers--;
                if(Amin2==1000000) {//numbers!=0
                    data[Amin1_num] = -data[Amin1_num];
                }
                else {
                    data[Amin1_num] = -data[Amin1_num];
                    data[Amin2_num] = -data[Amin2_num];
                }
                sum-=2;
            }
        }
        }
        long long res=0;
        for(int i=0;i<cur;i++){
            int ID=num[i];
            if(data[ID]==1) res+=a[ID];
            else if(data[ID]==-1) res+=b[ID];
        }
        printf("%I64d\n",res);
        for(int i=0;i<len;i++){
            if(data[i]==1) putchar('(');
            else if(data[i]==-1) putchar(')');
        }
        puts("");
    }
}