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最新推荐文章于 2022-02-07 13:24:58 发布

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最新推荐文章于 2022-02-07 13:24:58 发布

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分类专栏：题解

本文链接：https://blog.youkuaiyun.com/qq_37804582/article/details/83868355

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探讨了在给定括号序列中用最少成本替换问号以形成合法括号序列的算法。输入包括序列和替换成本，输出为最小总成本及相应序列。

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题目链接

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

Copy

(??)
1 2
2 8

output

Copy

4
()()

AC代码：

#include<cstdio>
using namespace std;
char e[4][4];
inline int check(char ch){
    int cnt = 0;
    bool flag1, flag2;
    for(int i = 0; i < 3; i++){
        flag1 = false, flag2 = false;
        for(int j = 0; j < 3; j++){
            if(e[i][j] != ch) flag1 = true;
            if(e[j][i] != ch) flag2 = true;
        }
        if(!flag1) cnt++;
        if(!flag2) cnt++;
    }
    flag1 = false, flag2 = false;
    for(int i = 0; i < 3; i++){
        if(e[i][i] != ch) flag1 = true;
        if(e[i][2 - i] != ch) flag2 = true;
    }
    if(!flag1) cnt++;
    if(!flag2) cnt++;
   // printf("%d\n",cnt);
    return cnt;
}
int main(){
    #ifdef ONLINE_JUDGE
    #else
        freopen("in.txt", "r", stdin);
    #endif // ONLINE_JUDGE
    for(int i = 0; i < 3; i++){
        scanf("%s", e[i]);
    }
    int a = 0, b = 0, c = 0;
    for(int i = 0; i < 3; i++){
        for(int j = 0; j < 3; j++){
            if(e[i][j] == 'X') a++;
            else if(e[i][j] == '0') b++;
            else if(e[i][j] == '.') c++;
        }
    }
    int ans1 = check('X');
    int ans2 = check('0');
    if(b > a || (a - b) > 1 || (ans1 && ans2)){
        printf("illegal\n");
        return 0;
    }
    if(ans1 >= 1 && ans2 == 0){
        if(a == b) printf("illegal");
        else printf("the first player won\n");
    }
    else if(ans1 == 0 && ans2 >= 1){
        if(a == b + 1) printf("illegal\n");
        else printf("the second player won\n");
    }
    else{
        if(a == b) printf("first\n");
        else if(a == b + 1){
            if(c == 0) printf("draw\n");
            else printf("second\n");
        }
        else printf("illegal\n");
    }
    return 0;
}