Mahjong Sorting

链接：https://zoj.pintia.cn/problem-sets/91827364500/problems/91827370268

Time Limit: 1000 msMemory Limit: 65536 KB

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 .

Output

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

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

Sample Output

2
4
7
25

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.

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=5e5+10;
long long mod=1e9;
char a[N][3];
int b[N],ans,flag,t,n,m;
void solve()
{
    if(flag==-1){
                        ans=m*3-n+1;
                    }
                    else if(flag==1){
                        if(a[2][0]=='C')ans=b[2]-1;
                        else if(a[2][0]=='B')ans=m+b[2]-1;
                        else ans=m*2+b[2]-1;
                    }
                    else if(n==2&&flag==2){
                        if(a[1][0]=='C')ans=m*3-b[1]+1;
                        else if(a[1][0]=='B')ans=m*2-b[1]+1;
                        else ans=m-b[1]+1;
                    }
                    else if(flag==n){
                        if(a[n-1][0]=='C')ans=m*3-b[n-1];
                        else if(a[n-1][0]=='B')ans=m*2-b[n-1];
                        else ans=m-b[n-1];
                    }
                    else {
                        if(flag==2){
                            if(a[1][0]==a[3][0])ans=b[3]-b[1];
                            else if(a[1][0]=='C'&&a[3][0]=='D')ans=m*2+b[3]-b[1];
                            else ans=m+b[3]-b[1];
                        }
                        else {
                            int x=flag;
                            if(a[x-1][0]==a[x+1][0])ans=b[x+1]-b[x-1]-1;
                            else if(a[x-1][0]=='C'&&a[x+1][0]=='D')ans=m*2+b[x+1]-b[x-1]-1;
                            else ans=m+b[x+1]-b[x-1]-1;
                        }
                    }
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&m);
        flag=-1;
        for(int i=1;i<=n;i++)
        {
            scanf("%s",a[i]);
            if(a[i][0]=='W')
            {
                flag=i;
            }
            else
            {
                scanf("%d",&b[i]);
            }
        }
        if(n==1){
            ans=m*3;
        }
        else
        {
            if(a[1][0]==a[2][0])
            {
                if(b[1]>b[2])
                {
                    ans=1;
                }
                else
                {
                    solve();
                }
            }
            else if(a[1][0]=='C')
            {
                if(a[2][0]=='W')
                {
                    if(n==2)
                    {
                        if(a[1][0]=='C')ans=m*3-b[1]+1;
                        else if(a[1][0]=='B')ans=m*2-b[1]+1;
                        else ans=m-b[1]+1;
                    }
                    else
                    {
                        if(a[3][0]=='C')
                        {
                            if(b[1]>b[3])
                                ans=1;
                            else
                            solve();
                        }
                        else
                            solve();
                    }
                }
                else
                {
                    solve();
                }
            }
            else if(a[1][0]=='B')
            {
                if(a[2][0]=='C')
                    ans=1;
                else if(a[2][0]=='W')
                {
                    if(n==2)
                    {
                        if(a[1][0]=='C')ans=m*3-b[1]+1;
                        else if(a[1][0]=='B')ans=m*2-b[1]+1;
                        else ans=m-b[1]+1;
                    }
                    else
                    {
                        if(a[3][0]=='C')
                            ans=1;
                        else if(a[3][0]=='B')
                        {
                            if(b[1]>b[3])
                                ans=1;
                            else
                            solve();
                        }
                        else
                            solve();
                    }
                }
                else
                {
                    solve();
                }
            }
            else
            {
                if(a[2][0]=='C'||a[2][0]=='B')
                    ans=1;
                else if(a[2][0]=='W')
                {
                    if(n==2)
                    {
                        if(a[1][0]=='C')ans=m*3-b[1]+1;
                        else if(a[1][0]=='B')ans=m*2-b[1]+1;
                        else ans=m-b[1]+1;
                    }
                    else
                    {
                        if(a[3][0]=='C'||a[3][0]=='B')
                            ans=1;
                        else if(a[3][0]=='D')
                        {
                            if(b[1]>b[3])
                                ans=1;
                            else
                            solve();
                        }
                        else
                            solve();
                    }
                }
                else
                {
                    solve();
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}