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
题意：
输入一个有括号组成的字符串，只可以改变左括号，算出最少要改变多少次才能使括号配对。
方法：
先输入一个字符串s，计算字符串长度K，定义一个栈st，i从0到K，如果s[i]为右括号，判断栈是否为空，若为空输出Impossible，若st.top()为对应的左括号则删除，否则记录sum++表示要修改的次数。
1 #include<cstdio> 2 #include<stack> 3 #include<string.h> 4 using namespace std; 5 int main() 6 { 7 char s[1000000+11]; 8 int k,sum; 9 while(scanf("%s",&s)!=EOF) 10 { 11 12 sum=0; 13 stack<char>st; 14 k=strlen(s); 15 for(int i = 0; i < k ; i++) 16 { 17 if(s[i] == '>') 18 { 19 if(st.empty()) 20 { 21 printf("Impossible "); 22 sum=-1; 23 break; 24 } 25 if(st.top() == '<') 26 { 27 st.pop(); 28 } 29 else 30 { 31 sum++; 32 st.pop(); 33 } 34 } 35 else if(s[i] == ']') 36 { 37 if(st.empty()) 38 { 39 printf("Impossible "); 40 sum=-1; 41 break; 42 } 43 if(st.top() == '[') 44 { 45 st.pop(); 46 } 47 else 48 { 49 sum++; 50 st.pop(); 51 } 52 } 53 else if(s[i] == ')') 54 { 55 if(st.empty()) 56 { 57 printf("Impossible "); 58 sum=-1; 59 break; 60 } 61 if(st.top() == '(') 62 { 63 st.pop(); 64 } 65 else 66 { 67 sum++; 68 st.pop(); 69 } 70 } 71 else if(s[i] == '}') 72 { 73 if(st.empty()) 74 { 75 printf("Impossible "); 76 sum=-1; 77 break; 78 } 79 if(st.top() == '{') 80 { 81 st.pop(); 82 } 83 else 84 { 85 sum++; 86 st.pop(); 87 } 88 } 89 else 90 st.push(s[i]); 91 } 92 if(sum != -1) 93 { 94 if(!st.empty()) 95 printf("Impossible "); 96 else 97 printf("%d ",sum); 98 } 99 } 100 101 }