ZOJ - 4034 Mahjong Sorting（模拟）（阅读理解题!!! ）_g - mahjong sorting zoj - 4034

 

DreamGrid has just found a set of Mahjong with  suited tiles and a White Dragon tile in his pocket. Each suited tile has a suit (Character, Bamboo or Dot) and a rank (ranging from 1 to ), and there is exactly one tile of each rank and suit combination.

Character tiles whose rank ranges from 1 to 9

Bamboo tiles whose rank ranges from 1 to 9

Dot tiles whose rank ranges from 1 to 9

White Dragon tile

As DreamGrid is bored, he decides to play with these tiles. He first selects one of the  suited tiles as the "lucky tile", then he picks  tiles from the set of  tiles and sorts these  tiles with the following rules:

• The "lucky tile", if contained in the  tiles, must be placed in the leftmost position.

• For two tiles  and  such that neither of them is the "lucky tile", if

  •  is a Character tile and  is a Bamboo tile, or

  •  is a Character tile and  is a Dot tile, or

  •  is a Bamboo tile and  is a Dot tile, or

  •  and  have the same suit and the rank of  is smaller than the rank of ,

  then  must be placed to the left of .

   

 

White Dragon tile is a special tile. If it's contained in the  tiles, it's considered as the original (not-lucky) version of the lucky tile during the sorting. For example, consider the following sorted tiles, where "3 Character" is selected as the lucky tile. White Dragon tile, in this case, is considered to be the original not-lucky version of "3 Character" and should be placed between "2 Character" and "4 Character".

As DreamGrid is quite forgetful, he immediately forgets what the lucky tile is after the sorting! Given  sorted tiles, please tell DreamGrid the number of possible lucky tiles.

Input

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

The first line contains two integers  and  (, ), indicating the number of sorted tiles and the maximum rank of suited tiles.

For the next  lines, the -th line describes the -th sorted tile counting from left to right. The line begins with a capital letter  (), indicating the suit of the -th tile:

• If , then an integer  () follows, indicating that it's a Character tile with rank ;

• If , then an integer  () follows, indicating that it's a Bamboo tile with rank ;

• If , then an integer  () follows, indicating that it's a Dot tile with rank ;

• If , then it's a White Drangon tile.

 

It's guaranteed that there exists at least one possible lucky tile, and the sum of  in all test cases doesn't exceed .

<h4< dd="">Output

For each test case output one line containing one integer, indicating the number of possible lucky tiles.

<h4< dd="">Sample Input

4
3 9
C 2
W
C 4
6 9
C 2
C 7
W
B 3
B 4
D 2
3 100
C 2
W
C 9
3 9
C 1
B 2
D 3

<h4< dd="">Sample Output

2
4
7
25

<h4< dd="">Hint

For the first sample, "2 Character" and "3 Character" are possible lucky tiles.

For the second sample, "8 Character", "9 Character", "1 Bamboo" and "2 Bamboo" are possible lucky tiles.

 

题意：

（翻译得想哭...）

（争取讲明白）

感谢老铁 shang Qin7~~（双击666）

 

 

麻将牌，规则如下：

有3m+1张牌，

m张“万”

m张“条”

m张“筒”

1张白板

 

在3m+1张牌中选n张牌，但在选之前先在3m张牌（除白板外）里挑选一张牌规定它为“幸运牌”

（先理解了这句话再继续往下看）

如果n张牌里包含这张“幸运牌”，那么这张“幸运牌”一定是在最左边的位置（第一张）；

如果n张牌里包含白板，那么这张白板相当于“不幸运”版本的“幸运牌”，（也就是只是和“幸运牌”的点数和种类一样，没有"幸运"这个性质。）比如题中给的例子，“三万”作为”幸运牌“，白板就是“不幸版本”的“三万”，所以放在“二万”和“四万”中间。

这么多相当于是说相同种类的牌一定是点数小的在点数大的牌的左边，种类从左到右依次是“万”“条”“筒”，也就是给出的n张牌是“一万”“两万”...“九万”“一条”“两条”...“九条”“一筒”“两筒”...“九筒”这个顺序。

 

下面给出你已经挑选出来的n张牌，问你可能的“幸运牌”数量是多少。（不是只有9种牌）

 

 

解题思路：

 

总结一下题意，“幸运牌”如果在这n张牌里那么一定是在n张牌的最左边，如果不在，那么n张牌里的白板会相当于（不幸版本）的“幸运牌”，需要放在合适的位置（题中所给的顺序），如果没有白板，只能n张牌里的第一张牌和n-1张牌之外的牌（即3*m-(n-1)）作为”幸运牌“了（这句话不好理解就先放这）。

 

先把给出的3m+1张牌都编上号便于判断与处理。

 

把所有情况列出来总的有以下几种：

 

1.当只有1张牌时，不管这一张牌是3m里的牌还是白板，都可以当作幸运牌，结果为3m；

2.当n>=1张牌时：

①当n>=2时当第一张牌比第二张牌的点数大并且第二张不是白板时，结果为1（第1张牌是幸运牌）

②当n张牌里没有白板时，结果为3*m-（n-1）（这里的1表示第一张牌，也就是n张牌里的第1张牌可以做幸运牌，剩下n-1张牌都不能做幸运牌，还有n张牌外的所有牌（除白板））

③当有白板时：

Ⅰ当白板是第一张牌时，结果为第二张牌的点数

Ⅱ当白板是第二张牌时，结果为第三张牌的点数-第一张牌的点数

Ⅲ当白板在其余位置时，结果为白板后第一张牌的点数-白班前一张牌的点数-1

（Ⅳ当白板在最后一位时，因为开始时处理了3m+1张牌的编号，所以如果白板在最后一位可以直接用Ⅲ来算）

 

AC代码:

#include<stdio.h>
#include<string.h>
 
int a[100005];
 
int main()
{
    int t,m,n,k,p;
    scanf("%d",&t);
    while(t--)
    {
        char str[3];
        scanf("%d%d",&n,&m);
        p = -1;
        a[n] = 3*m+1;                   //防止最后一个数是白板
        for(int i = 0; i < n; i++)
        {
            scanf("%s",str);
            if(str[0] == 'W')
                {
                    p = i;
                    continue;
                }
            scanf("%d",&k);
            if(str[0] == 'C')
                a[i] = k;
            else if(str[0] == 'B')
                a[i] = m+k;
            else
                a[i] = 2*m+k;
        }
        if(n == 1)
            printf("%d\n",3*m);              //只有一个数
        else if(a[0] > a[1] && p != 1)       //当第一个数比第二个数大时，并且第二个数不是白板，输出1
            printf("1\n");
        else if(p == -1)                     //没有白板的情况
            printf("%d\n",3*m-n+1);
        else
        {
            if(p == 0)                      //第一个数是白板的情况
                printf("%d\n",a[1]);
            else if(p == 1)                 //第二个数是白板的情况
                printf("%d\n",a[2]-a[0]);
            else                            //第p个数是白板的情况(p>1),等于其后一个数减去前一个数再减一
                printf("%d\n",a[p+1]-a[p-1]-1);
        }
    }
    return 0;//
}