poj 1068 Parencodings

Parencodings

Time Limit: 1000MS      Memory Limit: 10000K

Total Submissions: 25474 Accepted: 14988

Description

Let S = s1 s2…s2n be a well-formed string of parentheses. S can be encoded in two different ways:

q By an integer sequence P = p1 p2…pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).

q By an integer sequence W = w1 w2…wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).

Following is an example of the above encodings:

S       (((()()())))

P-sequence      4 5 6666

W-sequence      1 1 1456

Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.

Input

The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case is an integer n (1 <= n <= 20), and the second line is the P-sequence of a well-formed string. It contains n positive integers, separated with blanks, representing the P-sequence.

Output

The output file consists of exactly t lines corresponding to test cases. For each test case, the output line should contain n integers describing the W-sequence of the string corresponding to its given P-sequence.

Sample Input

2

6

4 5 6 6 6 6

9

4 6 6 6 6 8 9 9 9

Sample Output

1 1 1 4 5 6

1 1 2 4 5 1 1 3 9

题目大意：先给一个n，然后后面是n个数；

这个n个数，每个数分别代表当前有一个右括号，这个右括号前面有a[i]个左括号

求每个右括号与之相匹配的左括号之间有多少对成对的括号（包括他们自己）

#include <iostream>
#include <cstring>
#include <stack>

using namespace std;

bool vis[110];//用一个标记数组来标记一下右括号的位置

void Print(string s){
char str[110];//用一个一个数组来模拟栈
memset(str, false, sizeof(str));
int top = 0;
int len = s.length();
bool flag = true;
for (int i = 0; i < len; i++){
if (vis[i]){//当前位置为右括号时就搜索前面的，直到找到与之相匹配的就退出
//cout << top << endl;
int count1 = 0;
str[++top] = s[i];
for (int j = i - 1; j >= 0; j--){
if (s[j] == ')'){//遇到右括号就直接进栈
str[++top] = s[j];
}
else if (s[j] == '('){//遇到左括号就出栈，并记录一下个数
top--;
count1++;
if (top == 0){
if (flag){
cout << count1;
flag = false;
break;
}
else{
cout << ' ' << count1;
break;
}
}
}
}
}
}
cout << endl;
}

int main(){
int t;
cin >> t;
while (t--){
int n;
string s = "";
cin >> n;
int now;
int x;//输入的数代表当前位置前面一共有x个左括号，然后是一个右括号
cin >> x;
for (int i = 0; i < x; i++) s += '(';
now = x;
s += ')';
for (int i = 1; i < n; i++){
cin >> x;
for (int i = 0; i < x - now; i++) s += '(';
s += ')';
now = x;
}
memset(vis, false, sizeof(vis));
for (int i = 0; i < s.length(); i++){//标记右括号位置
if (s[i] == ')') vis[i] = true;
}
//      for (int i = 0; i < s.length(); i++)
//            (i == s.length() - 1) ? (cout << vis[i] << endl) : (cout << vis[i] << ' ');
Print(s);
}
return 0;
}