C. Replace To Make Regular Bracket Sequence

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本文介绍了一个算法问题，即如何通过最少的替换操作将一个由四种括号组成的字符串转化为合法的括号序列。文章提供了一段C语言代码实现，并详细解释了其工作原理。

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：

.。