Replace To Make Regular Bracket Sequence
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Status
Description
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.
Sample Input
Input
[<}){}
Output
2
Input
{()}[]
Output
0
Input
]]
Output
Impossible
算法的设计思想：
1）凡出现左括弧，则进栈；
2）凡出现右括弧，首先检查栈是否空
若栈空，则表明该“右括弧”多余，
否则
和栈顶元素
比较，
若相匹配，则“左括弧出栈” ，
否则表明不匹配。
3）表达式检验结束时，
若栈空，则表明表达式中匹配正确，
否则表明“左括弧”有余
代码：
#include<stdio.h>
#include<string.h>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;
char a[1000000+11];
int main()
{
int len;
while(scanf("%s",a)!=EOF)
{
stack<char>s;
int left=0,right=0,ans=0;
bool flag=true;
len=strlen(a);
for(int i=0;i<len;i++)
{
if(a[i]=='('||a[i]=='{'||a[i]=='['||a[i]=='<')
{
s.push(a[i]);
left++;
}
else if(a[i]==')')
{
right++;
if(s.empty())
{
flag=false;
break;
}
else
{
if(s.top()=='(')
s.pop();
else
{
ans++;
s.pop();
}
}
}
else if(a[i]=='}')
{
right++;
if(s.empty())
{
flag=false;
break;
}
else
{
if(s.top()=='{')
s.pop();
else
{
ans++;
s.pop();
}
}
}
else if(a[i]==']')
{
right++;
if(s.empty())
{
flag=false;
break;
}
else
{
if(s.top()=='[')
{
s.pop();
}
else
{
ans++;
s.pop();
}
}
}
else if(a[i]=='>')
{
right++;
if(s.empty())
{
flag=false;
break;
}
else
{
if(s.top()=='<')
{
s.pop();
}
else
{
ans++;
s.pop();
}
}
}
}
if(right!=left)
flag=false;
if(flag)
printf("%d\n",ans);
else
printf("Impossible\n");
}
return 0;
}
