Codeforces 612C Replace To Make Regular Bracket Sequence 【stack】
2016-07-31 20:22
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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.
Sample test(s)
input
output
input
output
input
output
题意 就是括号配对,不同的是不同括号也可以配对,但是有条件
题目条件已经很清晰
还是栈
<span style="font-size:18px;">#include<cstdio>
#include<string.h>
#include<stack>
using namespace std;
char s[1000000+11];
int main()
{
scanf("%s",s);
int len=strlen(s),sum=0;
stack<char>sta;
if(s[0]==']'||s[0]==')'||s[0]=='}'||s[0]=='>')//这样的就没办法配对
printf("Impossible\n");
else
{
int i;
for(i=0;i<len;++i)
{
if(s[i]=='['||s[i]=='('||s[i]=='{'||s[i]=='<')//压入栈中
sta.push(s[i]);
else if(!sta.empty())
{
if(s[i]==')'&&sta.top()!='('||s[i]=='}'&&sta.top()!='{'||s[i]=='>'&&sta.top()!='<'||s[i]==']'&&sta.top()!='[')
++sum;//可以配对但是不是同一种类型,那就需要变
sta.pop();//同时出栈
}
else
{
if(i==len-1)
break;
}
}
if(sta.empty())//如果全部可以配对就是空栈
{
printf("%d\n",sum);
}
else
{
printf("Impossible\n");
}
}
return 0;
}</span>
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.
Sample test(s)
input
[<}){}
output
2
input
{()}[]
output
0
input
]]
output
Impossible
题意 就是括号配对,不同的是不同括号也可以配对,但是有条件
题目条件已经很清晰
还是栈
<span style="font-size:18px;">#include<cstdio>
#include<string.h>
#include<stack>
using namespace std;
char s[1000000+11];
int main()
{
scanf("%s",s);
int len=strlen(s),sum=0;
stack<char>sta;
if(s[0]==']'||s[0]==')'||s[0]=='}'||s[0]=='>')//这样的就没办法配对
printf("Impossible\n");
else
{
int i;
for(i=0;i<len;++i)
{
if(s[i]=='['||s[i]=='('||s[i]=='{'||s[i]=='<')//压入栈中
sta.push(s[i]);
else if(!sta.empty())
{
if(s[i]==')'&&sta.top()!='('||s[i]=='}'&&sta.top()!='{'||s[i]=='>'&&sta.top()!='<'||s[i]==']'&&sta.top()!='[')
++sum;//可以配对但是不是同一种类型,那就需要变
sta.pop();//同时出栈
}
else
{
if(i==len-1)
break;
}
}
if(sta.empty())//如果全部可以配对就是空栈
{
printf("%d\n",sum);
}
else
{
printf("Impossible\n");
}
}
return 0;
}</span>
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