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hdoj Scaena Felix

时间:2015-09-26 22:35:42      阅读:180      评论:0      收藏:0      [点我收藏+]

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Scaena Felix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 101    Accepted Submission(s): 49


Problem Description
Given a parentheses sequence consist of ‘(‘ and ‘)‘, a modify can filp a parentheses, changing ‘(‘ to ‘)‘ or ‘)‘ to ‘(‘.

If we want every not empty <b>substring</b> of this parentheses sequence not to be "paren-matching", how many times at least to modify this parentheses sequence?

For example, "()","(())","()()" are "paren-matching" strings, but "((", ")(", "((()" are not.
 

 

Input
The first line of the input is a integer T, meaning that there are T test cases.

Every test cases contains a parentheses sequence S only consists of ‘(‘ and ‘)‘.

1|S|1,000.
 

 

Output
For every test case output the least number of modification.
 

 

Sample Input
3 () (((( (())
 

 

Sample Output
1 0 2
题解:水题。。。。匹配括号的个数。。。
代码:
 1 #include<stdio.h>
 2 #include<stack>
 3 using namespace std;
 4 const int MAXN=1100;
 5 char m[MAXN];
 6 int main(){
 7     int T;
 8     scanf("%d",&T);
 9     while(T--){
10         stack<char>st;
11         scanf("%s",m);
12         int k=0;
13         for(int i=0;m[i];i++){
14             if(m[i]==()st.push(m[i]);
15             else if(!st.empty())st.pop(),k++;
16         }
17         printf("%d\n",k);
18     }
19     return 0;
20 }

 

hdoj Scaena Felix

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原文地址:http://www.cnblogs.com/handsomecui/p/4841414.html

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