2015 Multi-University Training Contest 6 hdu 5357 Easy Sequence

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原文链接：http://www.cnblogs.com/crackpotisback/p/4742410.html

文章标签：

#java

本文介绍了一个字符串匹配问题的解决方法，使用栈来辅助计算有效子串的数量，并通过样例详细解释了计算过程。该问题涉及有效括号串的判断及计数。

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Easy Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 766 Accepted Submission(s): 208

Problem Description
soda has a string containing only two characters -- '(' and ')'. For every character in the string, soda wants to know the number of valid substrings which contain that character.

Note:
An empty string is valid. If S is valid, (S) is valid. If U,V are valid, UV is valid.

Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

A string s consisting of '(' or ')' $(1 \leq |s| \leq 10^6)$.

Output
For each test case, output an integer $m=\sum_{i=1}^{|s|}(i⋅ansi mod 1000000007)$, where ansi is the number of valid substrings which contain i-th character.

Sample Input
2
()()
((()))

Sample Output
20
42

Hint

For the second case, $ans = \{1, 2, 3, 3, 2, 1\}$, then $m=1 \cdot 1 + 2 \cdot 2 + 3 \cdot 3 + 4 \cdot 3 + 5 \cdot 2 + 6 \cdot 1 = 42$

Author
zimpha@zju

Source

2015 Multi-University Training Contest 6 1005

解题：栈

貌似还是有点不明白，代码先放着

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const LL mod = 1000000007;
 5 const int maxn = 2000010;
 6 char str[maxn];
 7 int stk[maxn],match[maxn],pre[maxn],a[maxn],b[maxn],top;
 8 LL ans[maxn];
 9 int main() {
10     int kase,n;
11     scanf("%d",&kase);
12     while(kase--) {
13         scanf("%s",str + 1);
14         top = 0;
15         n = strlen(str + 1);
16         for(int i = 1; i <= n; ++i) {
17             match[i] = pre[i] = 0;
18             if(str[i] == '(') stk[++top] = i;
19             else if(top) {
20                 match[stk[top]] = i;
21                 match[i] = stk[top];
22                 if(top > 1) pre[match[i]] = stk[top-1];
23                 stk[top--] = 0;
24             }
25         }
26         ans[0] = a[0] = b[n+1] = 0;
27         for(int i = 1; i <= n; i++)
28             a[i] = (str[i] == ')' && match[i])?(a[match[i] - 1] + 1):0;
29         for(int i = n; i >= 1; i--)
30             b[i] = (str[i] == '(' && match[i])?(b[i] = b[match[i] + 1] + 1):0;
31         for(int i = 1; i <= n; i++)
32             ans[i] = (str[i] == '(')?(ans[pre[i]] + ((LL)b[i]*a[match[i]] % mod) % mod):ans[match[i]];
33         LL ret = 0;
34         for(int i = 1; i <= n; ++i)
35             ret += ans[i]*i%mod;
36         printf("%I64d\n",ret);
37     }
38     return 0;
39 }