POJ 3991 Seinfeld

最新推荐文章于 2024-03-21 22:40:04 发布

转载最新推荐文章于 2024-03-21 22:40:04 发布 · 36 阅读

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原文链接：http://www.cnblogs.com/toyking/p/3797382.html

此博客讨论了如何通过替换操作将不稳定的括号字符串转换为稳定的字符串所需的最少步骤。涉及基本算法逻辑，适用于初学者理解字符串操作。

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Seinfeld

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 897		Accepted: 440

Description

I’m out of stories. For years I’ve been writing stories, some rather silly, just to make simple problems look difficult and complex problems look easy. But, alas, not for this one.
You’re given a non empty string made in its entirety from opening and closing braces. Your task is to find the minimum number of “operations” needed to make the string stable. The definition for being stable is as follows:

An empty string is stable.
If S is stable, then {S} is also stable.
If S and T are both stable, then ST (the concatenation of the two) is also stable.

All of these strings are stable: {}, {}{}, and {{}{}}; But none of these: }{, {{}{, nor {}{.
The only operation allowed on the string is to replace an opening brace with a closing brace, or visa-versa.

Input

Your program will be tested on one or more data sets. Each data set is described on a single line. The line is a non-empty string of opening and closing braces and nothing else. No string has more than 2000 braces. All sequences are of even length.
The last line of the input is made of one or more ’-’ (minus signs.)

Output

For each test case, print the following line:
k. N
Where k is the test case number (starting at one,) and N is the minimum number of operations needed to convert the given string into a balanced one.

Sample Input

}{
{}{}{}
{{{}
---

Sample Output

1. 2
2. 0
3. 1

Source

anarc 2009

这题很水，不涉及任何算法，只需把握这个规律：从左往右扫一遍，左括号一定比右括号多，或相等。如果不是把右括号变为左括号。

#include <iostream>
#include <string>
#include <cmath>
using namespace std;

string st;
int casen=0;

void computing(){
	if(st.length()<=0){
		cout<<casen<<". 0"<<endl;
		return;
	}
	int len=st.length(),c=0,ans=0;
	for(int i=0;i<len;i++){
		if(st[i]=='{') c++;
		if(st[i]=='}') c--;
		if(c<0){
			ans++;
			st[i]='{';
			i--;
			c++;
		}
	}
	cout<<casen<<". "<<ans+c/2<<endl;
}

int main(){
	while(getline(cin,st)){
		casen++;
		if(st.length()>0 && st[0]=='-') break;
		computing();
	}
	return 0;
}

转载于:https://www.cnblogs.com/toyking/p/3797382.html