【CodeForces】612C - Replace To Make Regular Bracket Sequence（栈，括号配对问题）
【CodeForces】612C - Replace To Make Regular Bracket Sequence（栈，括号配对问题）
C. Replace To Make Regular Bracket Sequence
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2,(s1)s2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string s RBS.
Input
The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.
Output
If it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Examples
input
[<}){}output
2input
{()}[]output
0input
]]output
Impossible栈括号配对的题，本来不难，刚开始还是WA了。后来添加了两个变量 left 和 right ，记录左右括号的数量，若不相等，则说明没有完成配对。
代码如下：
#include <cstdio>
#include <queue>
#include <stack>
#include <algorithm>
#include <cstring>
using namespace std;
char a[1000011];
int main()
{
int l;
int ans;
bool dot;
int left,right;
while (~scanf ("%s",a))
{
l=strlen(a);
stack<char>s;
dot=true;
ans=0;
int left=0,right=0;
for (int i=0;i<l;i++)
{
if (a[i]=='(' || a[i]=='<' || a[i]=='{' || a[i]=='[')
{
s.push(a[i]);
left++;
}
else if (a[i]==')')
{
right++;
if (s.empty())
{
dot=false;
break;
}
else if (s.top()=='(')
{
s.pop();
}
else
{
ans++;
s.pop();
}
}
else if (a[i]==']')
{
right++;
if (s.empty())
{
dot=false;
break;
}
else if (s.top()=='[')
{
s.pop();
}
else
{
ans++;
s.pop();
}
}
else if (a[i]=='}')
{
right++;
if (s.empty())
{
dot=false;
break;
}
else if (s.top()=='{')
{
s.pop();
}
else
{
ans++;
s.pop();
}
}
else if (a[i]=='>')
{
right++;
if (s.empty())
{
dot=false;
break;
}
else if (s.top()=='<')
{
s.pop();
}
else
{
ans++;
s.pop();
}
}
}
if (left!=right)
dot=false;
if (dot)
printf ("%d\n",ans);
else
printf ("Impossible\n");
}
return 0;
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