【BZOJ】3300: [USACO2011 Feb]Best Parenthesis（模拟）-优快云博客

本文介绍了一种计算平衡括号字符串得分的方法。通过深度优先搜索（DFS）算法，实现字符串匹配并计算得分。字符串“（）”得分为1，对于任意平衡字符串“A”，其得分为2*S(A)，若“A”，“B”得分分别为S(A)和S(B)，则“AB”得分为S(A)+S(B)。

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http://www.lydsy.com/JudgeOnline/problem.php?id=3300

这个细节太多QAQ

只要将所有的括号'('匹配到下一个')'然后dfs即可

简单吧，，，

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%lld", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr2(a, b, c) for1(i, 1, b) { for1(j, 1, c) cout << a[i][j]; cout << endl; }
#define printarr1(a, b) for1(i, 1, b) cout << a[i] << ' '; cout << endl
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int N=100005;
const long long MD=12345678910;
int n, q[N], top, inext[N];
long long dfs(int l, int r) {
	int rr=inext[l];
	long long ret=0;
	if(l!=rr-1) ret=(ret+((dfs(l+1, rr-1)<<1)%MD))%MD;
	else if(l==rr-1) ret=(ret+1)%MD;
	if(rr+1<=r) ret=(ret+(dfs(rr+1, r)%MD))%MD;
	return ret;
}

int main() {
	read(n);
	for1(i, 1, n) {
		int t=getint();
		if(!t) q[++top]=i;
		else inext[q[top--]]=i;
	}
	print(dfs(1, n));
	return 0;
}

Description

Recently, the cows have been competing with strings of balanced
parentheses and comparing them with each other to see who has the
best one.

Such strings are scored as follows (all strings are balanced): the
string "()" has score 1; if "A" has score s(A) then "(A)" has score
2*s(A); and if "A" and "B" have scores s(A) and s(B), respectively,
then "AB" has score s(A)+s(B). For example, s("(())()") =
s("(())")+s("()") = 2*s("()")+1 = 2*1+1 = 3.

Bessie wants to beat all of her fellow cows, so she needs to calculate
the score of some strings. Given a string of balanced parentheses
of length N (2 <= N <= 100,000), help Bessie compute its score.

计算“平衡字符串”的分数，“平衡字符串”是指由相同数量的‘（’和‘）’组成，
且以‘（’开头，以‘）’结尾的字符串。
计算规则：
字符串“（）”的得分是1.
如果，平衡字符串“A”的得分是是S(A)，那么字符串“（A）”得分是2*S(A) ；
如果，“A”，“B” 得分分别是S(A)和S(B)，那么平衡字符串“AB”得分为S(A)+S(B)
例如：s("(())()") =s("(())")+s("()") = 2*s("()")+1 = 2*1+1 = 3.

Input

* Line 1: A single integer: N

* Lines 2..N + 1: Line i+1 will contain 1 integer: 0 if the ith
character of the string is '(', and 1 if the ith character of
the string is ')'
第1行：N，平衡字符串长度
第2至N+1行：Linei+1 整数0或1,0代表字符‘（’，1代表‘）’