hdoj 5479 Scaena Felix 【Stack】

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本文探讨了如何计算最少修改次数使括号序列中每个非空子串都不成为正确匹配的括号序列的问题。介绍了问题背景及算法实现，并提供了一段C++代码作为解答示例。

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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;
}