• 2018 CCPC秦皇岛 C题 Crusader Quest


    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  () 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 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  (about 50), indicating the number of test cases. For each test case:

    The first line contains a string  () 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.

    <h4< dd="">Output

    For each test case, output an integer denoting the maximum number of times DreamGrid can trigger the super skill.

    <h4< dd="">Sample Input

    7
    gggaaaooo
    aaoogggoa
    googgaaao
    agogaooag
    goooggaaa
    gogogoaaa
    gaogaogao

    <h4< dd="">Sample Output

    3
    3
    2
    1
    3
    2
    1

    <h4< dd="">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.

    题解:暴力即可;

     1 #include <bits/stdc++.h>
     2 
     3 using namespace std;
     4 int cnt[26];
     5 int l[26], r[26];
     6 char s[15];
     7 char id[6][3] = {{'a','o','g'},
     8                  {'a','g','o'},
     9                  {'o','a','g'},
    10                  {'o','g','a'},
    11                  {'g','a','o'},
    12                  {'g','o','a'} };
    13 
    14 int check(char m[3])
    15 {
    16     int num = 0;
    17     cnt['a'-'a'] = cnt['o' - 'a'] = cnt['g' - 'a'] = 3;
    18     for(int i = 0; i < 3; i++)
    19     {
    20         if(cnt[m[i] - 'a'] == 3)
    21         {
    22             num++;
    23             for(int j = l[m[i] - 'a']; j <= r[m[i] - 'a']; j++)
    24             {
    25                 cnt[s[j] - 'a']--;
    26             }
    27         }
    28     }
    29     return num;
    30 }
    31 
    32 
    33 int main()
    34 {
    35     int n;
    36     scanf("%d", &n);
    37     while(n--)
    38     {
    39         scanf("%s", s);
    40         l['a'-'a'] = l['o' - 'a'] = l['g' - 'a'] = 100;
    41         r['a'-'a'] = r['o' - 'a'] = r['g' - 'a'] = -1;
    42         for(int i = 0; i < 9; i++)
    43         {
    44             l[s[i] - 'a'] = min(l[s[i] - 'a'], i);
    45             r[s[i] - 'a'] = max(r[s[i] - 'a'], i);
    46         }
    47         int ans = 0;
    48         for(int i = 0; i < 6; i++)
    49             ans = max(ans, check(id[i]));
    50         printf("%d
    ", ans);
    51     }
    52     return 0;
    53 }
    View Code

      

  • 相关阅读:
    ajax traditional
    阿里云OSS NET SDK 引用示范程序
    js对象的两种写法
    BZOJ NOIP提高组十连测第一场
    ikbc 时光机 F87 Ctrl 失灵 解决办法
    【读书笔记】阅读的危险
    51nod 1118 机器人走方格 解题思路:动态规划 & 1119 机器人走方格 V2 解题思路:根据杨辉三角转化问题为组合数和求逆元问题
    【算法】求逆元模板
    【复习资料】软件工程之快速原型模型
    VirtualBox安装linux mint教程
  • 原文地址:https://www.cnblogs.com/csushl/p/9787680.html
Copyright © 2020-2023  润新知