题目传送门
1 /*
2 题意：求最长括号匹配的长度和它的个数
3 贪心+stack：用栈存放最近的左括号的位置，若是有右括号匹配，则记录它们的长度，更新最大值，可以在O (n)解决
4 详细解释：http://blog.csdn.net/taoxin52/article/details/26012167
5 */
6 #include <cstdio>
7 #include <algorithm>
8 #include <cstring>
9 #include <cmath>
10 #include <stack>
11 using namespace std;
12
13 const int MAXN = 1e6 + 10;
14 const int INF = 0x3f3f3f3f;
15 char s[MAXN];
16 int len[MAXN];
17 stack<int> S;
18
19 int main(void) //Codeforces Beta Round #5 C. Longest Regular Bracket Sequence
20 {
21 // freopen ("C.in", "r", stdin);
22
23 while (scanf ("%s", s) == 1)
24 {
25 while (!S.empty ()) S.pop ();
26 memset (len, 0, sizeof (len));
27
28 int mx = 0; int cnt = 0;
29 for (int i=0; s[i]; ++i)
30 {
31 if (s[i] == '(') S.push (i);
32 else
33 {
34 if (!S.empty ())
35 {
36 int p = S.top (); S.pop ();
37 len[i] = (p > 0 ? len[p-1] : 0) + (i - p + 1);
38 }
39 }
40 if (len[i] > mx) {mx = len[i]; cnt = 1;}
41 else if (len[i] == mx) {cnt++;}
42 }
43
44 if (mx == 0) puts ("0 1");
45 else printf ("%d %d
", mx, cnt);
46 }
47
48 return 0;
49 }