Codeforces 612C Replace To Make Regular Bracket Sequence【栈】
2016-07-26 22:45
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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.
Examples
input
output
input
output
input
output
这是栈的括号配对问题,就像南阳OJ的题型,但是不同的是这道题中,不同类型的括号还可以互换,而且要是可以全部出栈,输出替换的次数。
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.
Examples
input
[<}){}
output
2
input
{()}[]
output
0
input
]]
output
Impossible
这是栈的括号配对问题,就像南阳OJ的题型,但是不同的是这道题中,不同类型的括号还可以互换,而且要是可以全部出栈,输出替换的次数。
#include <cmath> #include <queue> #include <stack> #include <cstdio> #include <cstring> #include <algorithm> #define MAX_N 1000005 using namespace std; const int INF = 0xffff; char s[MAX_N]; //用数组将每种括号存储下来,对于判断括号的类型就可以借助数组下标,比较方便 char c[] = {'<', '>', '(', ')', '{', '}', '[', ']'}; int main() { stack<char> stk; while (scanf("%s", s) != EOF) { while (!stk.empty()) stk.pop(); int ans = 0; int len = strlen(s), k, m, n; for (int i = 0; i < len; i++) { if (!stk.empty()) { //查找字符在数组中的下标,确定字符的类型 for (k = 0; k < 8; k++) { if (stk.top() == c[k]) m = k; if (s[i] == c[k]) n = k; } //类型不同并且出栈元素为左边的括号才可以出栈 if (m%2 != n%2 && !(m%2)) { if (m != n - 1) ans++; stk.pop(); } else stk.push(s[i]); } else stk.push(s[i]); } if (stk.empty()) printf("%d\n", ans); else printf("Impossible\n"); } return 0; }
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