UVA13029 Emoticons【文本处理】

Nowadays emoticon has become an art. People are no longer limited to simple ones like

‘:-)’, ‘:-(’,‘:-P’, etc. They use ‘>:O’, ‘~_~’, ‘=^_^=’

and so on. Recently I came across ‘^_^’ and it looks kind of cute to me. Given a string S consisting of only ‘_’s and ‘^’s, I was wondering what is the maximum number of disjoint subsequences of “^_^” (quote for beauty) in the string S.
For example, if S = “^^__^^” then the answer is 2. However, for S = “_^^” the answer is 0.

Input
Input starts with a positive integer T (≤ 5, 000), denoting the number of test cases. Hence follows T test cases. Each case consists of a single string made of only ‘^’ and ‘_’. The length of the strings would be at most 100,000 and the sum of lengths of the strings will be 2,100,000 at most.
Output
For each test case, print the case number followed by the answer.
Hint:
• S[1…n] means S is a string of length n and it is 1-indexed.
• Si means i’th character of S.
• A string S[1…n] is a subsequence of another string T[1…m], if we can find: (t1, t2, . . . , tn) such that, S[i] = T[ti] for 1 ≤ i ≤ n and 1 ≤ t1 < t2 < . . . < tn ≤ m. For example, ‘abc’ is a subsequence of ‘aabbcc’ but not of ‘bca’.
• Two subsequences are disjoint if same character (position matters) is not used in both of the subsequences. For example, let S = ‘abca’. ‘ab’ and ‘ca’ are two disjoint subsequences of S. However, if S = ‘abc’ then ‘ab’ and ‘ac’ are not disjoint subsequences. In both of these examples the subsequences are unique. However, for S = ‘aabb’ let’s form two subsequences S1S3 and S2S4 (both are ‘ab’), both of these are disjoint. But if we have chosen S1S3 and S1S4 then they would not be disjoint.
Sample Input

5
_^^_^^_
^__^__^
______
^^__^^
^_^^_^

Sample Output
Case 1: 1
Case 2: 1
Case 3: 0
Case 4: 2
Case 5: 2

问题链接：UVA13029 Emoticons
问题简述：（略）
问题分析：文本处理问题，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA13029 Emoticons */#include <bits/stdc++.h>using namespace std;const int N = 100000 + 1;
char s[N];int main()
{int t;scanf("%d", &t);for (int k = 1; k <= t; k++) {scanf("%s", s);int ans = 0, h1 = 0, h2 = 0, h3 = 0;for (int i = 0; s[i]; i++) {if (s[i] == '^') {if (h2) h2--, ans++;else if (h3 && ans) h2++, h3--;else h1++;} else if(s[i] == '_') {if (h1) h1--, h2++;else if (ans > h3) h3++;}}printf("Case %d: %d\n", k, ans);}return 0;
}

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