POJ 1068 Parencodings

Parencodings

Time Limit: 1000MS	Memory Limit: 10000K
Total Submissions: 27370	Accepted: 16089

Description
Let S = s1s2...s2n be a well-formed string of parentheses. S can be encoded in twodifferent ways: 

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

q By an integer sequence W = w1 w2...wn where for each right parenthesis, say ain S, we associate an integer which is the number of right parentheses countingfrom 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-sequenceof the same string. 
Input
The first line ofthe input contains a single integer t (1 <= t <= 10), the number of testcases, followed by the input data for each test case. The first line of eachtest case is an integer n (1 <= n <= 20), and the second line is theP-sequence
of a well-formed string. It contains n positive integers, separatedwith blanks, representing the P-sequence.
Output
The output fileconsists of exactly t lines corresponding to test cases. For each test case,the output line should contain n integers describing the W-sequence of thestring 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
 
模拟方法，首先构造字符串：
int produceS(int
n, int*
Psequence, char*
tmpc) {
       int k = 0;
       for (int i = 1; i <=
n; i++) {
              for (int j = 0; j<Psequence[i] -
Psequence[i - 1]; j++)
                     tmpc[k++] =
'(';
              tmpc[k++] =
')';
       }
       return k;
}
 

然后使用堆统计匹配的括号：

void solve(char*
tmpc, int
length) {
       stack<char> cstack;
       for (int i = 0; i<length; i++) {
              if (tmpc[i] ==
'(') { cstack.push('(');
continue; }
              int ans = 1;
              while(!cstack.empty() && cstack.top() !=
'(')
              {
                     ans +=cstack.top() -
'0';
                     cstack.pop();
              }
              cstack.pop();
              cstack.push('0' + ans);
              cout << ans;
              string tmp = i !=
length - 1 ? " " :
"";
              cout << tmp;
       }
       cout << endl;
}

 

 

完整代码如下：

#include
<cstdio>
#include
<cstring>
#include
<iostream>
#include
<vector>
#include
<stack>
#include
<cstring>
#include
<string>
using namespace std;
 
//poj 1068
 
int produceS(int
n, int*
Psequence, char*
tmpc) {
       int k = 0;
       for (int i = 1; i <=
n; i++) {
              for (int j = 0; j<Psequence[i] -
Psequence[i - 1]; j++)
                     tmpc[k++] =
'(';
              tmpc[k++] =
')';
       }
       return k;
}
void solve(char*
tmpc, int
length) {
       stack<char> cstack;
       for (int i = 0; i<length; i++) {
              if (tmpc[i] ==
'(') { cstack.push('(');
continue; }
              int ans = 1;
              while(!cstack.empty() && cstack.top() !=
'(')
              {
                     ans +=cstack.top() -
'0';
                     cstack.pop();
              }
              cstack.pop();
              cstack.push('0' + ans);
              cout << ans;
              string tmp = i !=
length - 1 ? " " :
"";
              cout << tmp;
       }
       cout << endl;
}
int main(int
argc, char
const *argv[])
{
       //freopen("in.txt","r", stdin);
       int caseTime;
       int n;
       int Psequence[50] ={ 0 };
       char tmpc[100];
       scanf("%d",&caseTime);
 
       for (int i = 0;i<caseTime; i++) {
              scanf("%d", &n);
              for (int j = 1; j <=n; j++)
                     scanf("%d",&Psequence[j]);
              int length =produceS(n, Psequence, tmpc);
              solve(tmpc, length);
       }
       return 0;
}