C. Replace To Make Regular Bracket Sequence

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.#include<stdio.h>
#include<stack>
#include<map>
using namespace std;
int main(void) {
int count = 0;
char ch;
stack<char>s;
map<char, int>a;
char pre[4] = { '<','[','{','(' }, suf[4] = { '>',']','}',')' };
a['<'] = a['>'] = 0; a['['] = a[']'] = 1; a['{'] = a['}'] = 2; a['('] = a[')'] = 3;
while ((ch = getchar()) != '\n') {
if (ch == '<' || ch == '[' || ch == '{' || ch == '(')
s.push(ch);
else {
if (s.empty()) { printf("Impossible\n"); return 0; }//全部为右括弧的极端情况
if (a[s.top()] != a[ch]) {
s.pop(); count++;
}
else s.pop();
}
}
if (s.empty())
printf("%d\n", count);
else
printf("Impossible\n");
return 0;
}


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.

题目原址：http://codeforces.com/problemset/problem/612/C

思路：每次碰到左括弧就压入栈，碰到右括弧则出栈，若匹配则弹出，不匹配则对修改次数+1。若输入终止了，栈中仍有元素，说明栈里有未匹配的左括弧。所以可用栈是否为空判断是否输出”impossible“。需要注意的一点是全部为右括弧的极端情况。

代码如下：

.。