hdoj 5479 Scaena Felix 【Stack】

题目链接：hdoj 5479 Scaena Felix

Scaena Felix

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 763 Accepted Submission(s): 325

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 substring 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

题意：就是求匹配括号数目

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <cmath>
#include <stack>
#define fi first
#define se second
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MOD = 1e9 + 7;
const int MAXN = 2*1e5 + 10;
void add(LL &x, LL y) { x += y; x %= MOD; }
char str[1010];
int main()
{
int t; scanf("%d", &t);
while(t--) {
scanf("%s", str);
int len = strlen(str); int ans = 0;
stack<int> S;
for(int i = 0; i < len; i++) {
if(str[i] == '(') {
S.push(1);
}
else {
if(!S.empty()) {
ans++;
S.pop();
}
}
}
printf("%d\n", ans);
}
return 0;
}