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.
2 6 4 5 6 6 6 6 9 4 6 6 6 6 8 9 9 9Sample Output
1 1 1 4 5 6
1 1 2 4 5 1 1 3 9
解題思路:
其實這是一道模擬題,像這種括號匹配問題,我覺得經常會用到堆,主要就是把各種情況都考慮到,就沒有什麼問題了,很早之前做的了
代碼:
#include<iostream>
#include<stack>
#include<cstring>
using namespace std;
int p[21];
int w[21];
char s[21];
int n;
int p_s()
{
int help;
for(int i = 0;i<n;i++)
{
if(i==0)
{
for(int j = 0;j<p[0];j++)
s[j] = '(';
help = p[0];
s[help] = ')';
}
else if((p[i]-p[i-1])!=0)
{
int help2 = p[i]-p[i-1];
for(int j = 0;j<help2;j++)
{
s[help+1+j] = '(';
}
help = help+help2+1;
s[help] = ')';
}
else if((p[i]-p[i-1])==0)
{
help++;
s[help] = ')';
}
}
//for(int i = 0;i<=help;i++)
// cout<<s[i];
//cout<<endl;
return help;
}
int main()
{
int times ;
cin>>times;
while(times--)
{
int qingkuang = 0;
int m = 0;
cin>>n;
for(int i = 0;i<n;i++)
cin>>p[i];
int big = p_s();
int jilu[200];
memset(jilu,0,sizeof(jilu));
stack<int>sta;
for(int i = 0;i<=big;i++)
{
if(s[i]=='(')
{
sta.push(i);
}
else if(s[i]==')')
{
if(s[i-1]=='(')
{
jilu[i] = 1;
sta.pop();
}
else
{
int x = sta.top();
jilu[i] = (i-x+1)/2;
sta.pop();
}
}
}
for(int i = 0;i<=big;i++)
{
if(jilu[i]!=0)
cout<<jilu[i]<<" ";
}
cout<<endl;
}
}