ZOJ - 3983

Crusaders Quest is an interesting mobile game. A mysterious witch has brought great darkness to the game world, and the only hope for your kingdom is to save the Goddesses so that they can unleash their power to fight against the witch.
In order to save the game world, you need to choose three heroes to fight for victory and use their skills wisely. Nine skill blocks of three different types (three blocks per type) will be presented at the bottom of the screen. If kkk (k≥1k \ge 1k≥1) consecutive blocks are of the same type, you can tap on them and eliminate them, thus triggering the powerful skill they represent. After the elimination, the blocks to their left will be connected with the blocks to their right. Moreover, if k=3k=3k=3 consecutive blocks of the same type are eliminated, the powerful skill they unleash will be upgraded to a super skill, which is the most powerful skill of all.
DreamGrid is a newbie in this game, and he wants to trigger the super skill as many times as he can. Given nine skill blocks satisfying the description above, please help DreamGrid calculate the maximum number of times he can trigger the super skill.
Input
There are multiple test cases. The first line of input contains an integer TTT (about 50), indicating the number of test cases. For each test case:
The first line contains a string sss (∣s∣=9|s| = 9∣s∣=9) consisting of three ‘g’s, three ‘a’s and three ‘o’s, representing the nine skill blocks of three different types. Each type of character represents one type of skill block.
Output
For each test case, output an integer denoting the maximum number of times DreamGrid can trigger the super skill.
Sample Input
7
gggaaaooo
aaoogggoa
googgaaao
agogaooag
goooggaaa
gogogoaaa
gaogaogao
Sample Output
3
3
2
1
3
2
1
Hint
For the first sample test case, DreamGrid can first eliminate “aaa” (one super skill triggered), thus changing the skill blocks to “gggooo”. He can then eliminate “ggg” (another super skill triggered) and finally eliminate “ooo” (a third super skill triggered). So the answer is 3.
For the second sample test case, DreamGrid can first eliminate “ggg” (one super skill triggered), thus changing the skill blocks to “aaoooa”. He can then eliminate “ooo” (another super skill triggered) and finally eliminate “aaa” (a third super skill triggered). So the answer is also 3.
For the third sample test case, DreamGrid can first eliminate “aaa” (one super skill triggered), thus changing the skill blocks to “googgo”. He can then eliminate “oo” to obtain “gggo”, and eliminate “ggg” (another super skill triggered) to obtain “o”. So the answer is 2. It is easy to prove that he cannot trigger the super skill three times under this arrangement of skill blocks.

先贪心将原字符串暴力消解。
如果全部消掉了，答案自然是3。暴力消解不存在消两对。
如果消去一对，那剩下的把其中一种消掉，留剩下一种凑成一次，此时答案是2。
如果暴力消解时一次都没消去，那答案不是2就是1。如果任意的字母的最左端到最右端的区间内有其他两个字母（要消去这两个字母才能使这个字母的两端连接起来，如果三种字母都这样的话，那就只能消去其他两种才能使得其中一种连接起来了），答案就是1，否则是2。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <unordered_set>
#include <algorithm>
using namespace std;
#define scanf scanf_s

string s;


int main() {
	ios::sync_with_stdio(false);
	int T;
	cin >> T;
	while (T--) {
		int Ans = 0;
		cin >> s;
		string temp = s;
		for (int i = 0; i < s.size(); ++i) {
			if (i + 2 < 9 && s[i] == s[i + 1] && s[i + 1] == s[i + 2]) {
				++Ans;
				s.erase(i, 3);
				i = max(i - 3, -1);
			}
		}

		if (Ans == 1) {
			Ans = 2;
		}
		else if (Ans == 0) {
			auto Find = [&temp](const char ch)->bool {
				int Left = -1, Right = -1;
				for (int i = 0; i < 9; ++i) {
					if (temp[i] == ch) {
						if (Left == -1) {
							Left = i;
						}
						else {
							Right = i;
						}
					}
				}
				char other = 0;
				for (int i = Left; i <= Right; ++i) {
					if (temp[i] != ch) {
						if (other == 0) {
							other = temp[i];
						}
						else if (temp[i] != other) {
							return true;
						}
					}
				}
				return false;
			};
			if (Find('a') && Find('g') && Find('o')) {
				Ans = 1;
			}
			else {
				Ans = 2;
			}
		}
		cout << Ans << endl;
	}

}