Codeforces 5C.Longest Regular Bracket Sequence(最长连续合法括号匹配[dp + stack])

Longest Regular Bracket Sequence

This is yet another problem dealing with regular bracket sequences.

We should remind you that a bracket sequence is called regular, if by inserting «+» and «1» into it we can get a correct mathematical expression. For example, sequences «(())()», «()» and «(()(()))» are regular, while «)(», «(()» and «(()))(» are not.

You are given a string of «(» and «)» characters. You are to find its longest substring that is a regular bracket sequence. You are to find the number of such substrings as well.

Input
The first line of the input file contains a non-empty string, consisting of «(» and «)» characters. Its length does not exceed 106.

Output
Print the length of the longest substring that is a regular bracket sequence, and the number of such substrings. If there are no such substrings, write the only line containing “0 1”.

Examples
Input
)((())))(()())
Output
6 2
Input
))(
Output
0 1

题意： 求最长连续合法括号匹配的长度及数量
思路： 偷的 首先是用stack保存每个 ‘(’ 的位置，也就是正常的括号匹配。
dp[i] 表示 i 位置以 ‘)’ 结尾 连续的合法括号匹配的长度。当 str[i] == ‘)’ 时，有
dp[i] = dp[st.top() - 1] + i - st.top() + 1
最后扫一遍 dp 数组求最大长度和数量即可。

Code：

#include<bits/stdc++.h>
#define debug(x) cout << "[" << #x <<": " << (x) <<"]"<< endl
#define pii pair<int,int>
#define clr(a,b) memset((a),b,sizeof(a))
#define rep(i,a,b) for(int i = a;i < b;i ++)
#define pb push_back
#define MP make_pair
#define LL long long
#define ull unsigned LL
#define ls i << 1
#define rs (i << 1) + 1
#define fi first
#define se second
#define ptch putchar
#define CLR(a) while(!(a).empty()) a.pop()

using namespace std;
inline LL read() {
    LL s = 0,w = 1;
    char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == '-') w = -1;
        ch = getchar();
    }
    while(isdigit(ch))
        s = s * 10 + ch - '0',ch = getchar();
    return s * w;
}
inline void write(LL x) {
    if(x < 0)
        putchar('-'), x = -x;
    if(x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}

const int maxn = 1e6 + 10;
char str[maxn];
int dp[maxn];
stack<int>st;

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    while(~scanf("%s",str + 1)){
        CLR(st); clr(dp,0);
        int n = strlen(str + 1);
        for(int i = 1;i <= n;++ i){
            if(str[i] == '(')
                st.push(i);
            else {
                if(st.empty()) continue;
                int top = st.top(); st.pop();
                dp[i] = dp[top - 1] + i - top + 1;
            }
        }
        int maxx = 0,num = 0;
        for(int i = 1;i <= n;++ i){
            if(dp[i] > maxx){
                maxx = dp[i];
                num = 1;
            }
            else if(dp[i] == maxx)
                ++ num;
        }
        write(maxx); ptch(' ');
        write(maxx == 0 ? 1 : num); ptch('\n');
    }
    return 0;
}