3300: [USACO2011 Feb]Best Parenthesis
Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 89 Solved: 42
[Submit][Status]
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‘）’
Output
* Line 1: The score of the string. Since this number can get quite
large, output the score modulo 12345678910.
计算字符串得分，结果对12345678910取模
Sample Input
6
0
0
1
1
0
1
INPUT DETAILS:
This corresponds to the string "(())()".
Sample Output
3
HINT
Source
题解：
题意不明。。。没有说清每个左括号能不能都匹配到右括号。。。导致我想了半天。。。
本来想在栈里面直接算出答案来的，结果发现我算不出来。。。
然后去写dfs，发现还是很好写的。。。
代码：
1 #include<cstdio>
2 #include<cstdlib>
3 #include<cmath>
4 #include<cstring>
5 #include<algorithm>
6 #include<iostream>
7 #include<vector>
8 #include<map>
9 #include<set>
10 #include<queue>
11 #include<string>
12 #define inf 1000000000
13 #define maxn 100000+1000
14 #define maxm 500+100
15 #define eps 1e-10
16 #define ll long long
17 #define pa pair<int,int>
18 #define for0(i,n) for(int i=0;i<=(n);i++)
19 #define for1(i,n) for(int i=1;i<=(n);i++)
20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
21 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
22 #define mod 12345678910
23 using namespace std;
24 inline int read()
25 {
26 int x=0,f=1;char ch=getchar();
27 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
28 while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
29 return x*f;
30 }
31 int p[maxn],n,top,sta[maxn];
32 inline ll dfs(int l,int r)
33 {
34 if(r==l+1)return 1;
35 ll tmp=0;
36 for(int i=l+1;i<r;i=p[i]+1)tmp+=2*dfs(i,p[i]),tmp%=mod;
37 return tmp;
38 }
39 int main()
40 {
41 freopen("input.txt","r",stdin);
42 freopen("output.txt","w",stdout);
43 n=read();
44 for1(i,n)
45 {
46 int x=read();
47 if(!x)sta[++top]=i;
48 else p[sta[top--]]=i;
49 }
50 ll ans=0;
51 for(int i=1;i<n;i=p[i]+1)ans+=dfs(i,p[i]),ans%=mod;
52 printf("%lld\n",ans);
53 return 0;
54 }
挖个坑，以后看看能不能想出在栈里直接搞定的方法。