Help Me with the Game(imitate)

Help Me with the Game
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 3630   Accepted: 2319

Description

Your task is to read a picture of a chessboard position and print it in the chess notation.

Input

The input consists of an ASCII-art picture of a chessboard with chess pieces on positions described by the input. The pieces of the white player are shown in upper-case letters, while the black player's pieces are lower-case letters. The letters are one of "K" (King), "Q" (Queen), "R" (Rook), "B" (Bishop), "N" (Knight), or "P" (Pawn). The chessboard outline is made of plus ("+"), minus ("-"), and pipe ("|") characters. The black fields are filled with colons (":"), white fields with dots (".").

Output

The output consists of two lines. The first line consists of the string "White: ", followed by the description of positions of the pieces of the white player. The second line consists of the string "Black: ", followed by the description of positions of the pieces of the black player. 

The description of the position of the pieces is a comma-separated list of terms describing the pieces of the appropriate player. The description of a piece consists of a single upper-case letter that denotes the type of the piece (except for pawns, for that this identifier is omitted). This letter is immediatelly followed by the position of the piece in the standard chess notation -- a lower-case letter between "a" and "h" that determines the column ("a" is the leftmost column in the input) and a single digit between 1 and 8 that determines the row (8 is the first row in the input). 

The pieces in the description must appear in the following order: King("K"), Queens ("Q"), Rooks ("R"), Bishops ("B"), Knights ("N"), and pawns. Note that the numbers of pieces may differ from the initial position because of capturing the pieces and the promotions of pawns. In case two pieces of the same type appear in the input, the piece with the smaller row number must be described before the other one if the pieces are white, and the one with the larger row number must be described first if the pieces are black. If two pieces of the same type appear in the same row, the one with the smaller column letter must appear first.

Sample Input

+---+---+---+---+---+---+---+---+
|.r.|:::|.b.|:q:|.k.|:::|.n.|:r:|
+---+---+---+---+---+---+---+---+
|:p:|.p.|:p:|.p.|:p:|.p.|:::|.p.|
+---+---+---+---+---+---+---+---+
|...|:::|.n.|:::|...|:::|...|:p:|
+---+---+---+---+---+---+---+---+
|:::|...|:::|...|:::|...|:::|...|
+---+---+---+---+---+---+---+---+
|...|:::|...|:::|.P.|:::|...|:::|
+---+---+---+---+---+---+---+---+
|:P:|...|:::|...|:::|...|:::|...|
+---+---+---+---+---+---+---+---+
|.P.|:::|.P.|:P:|...|:P:|.P.|:P:|
+---+---+---+---+---+---+---+---+
|:R:|.N.|:B:|.Q.|:K:|.B.|:::|.R.|
+---+---+---+---+---+---+---+---+

Sample Output

White: Ke1,Qd1,Ra1,Rh1,Bc1,Bf1,Nb1,a2,c2,d2,f2,g2,h2,a3,e4
Black: Ke8,Qd8,Ra8,Rh8,Bc8,Ng8,Nc6,a7,b7,c7,d7,e7,f7,h7,h6

Source

 
  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<ctype.h>
  4 #include<math.h>
  5 #include<algorithm>
  6 using namespace std;
  7 char st[40] , chess[10][40] ;
  8 int a[9][9] ;
  9 
 10 struct white
 11 {
 12     char ch , row , col;
 13     int pow ;
 14 }wh[100];
 15 
 16 struct black
 17 {
 18     char ch , row , col ;
 19     int pow ;
 20 }bl[100];
 21 
 22 bool cmp2 (white a , white b)
 23 {
 24     if (a.pow < b.pow)
 25         return 1 ;
 26     if (a.pow > b.pow)
 27         return 0 ;
 28     if (a.pow == b.pow ) {
 29         if (a.row == b.row)
 30             return a.col < b.col ;
 31         else
 32             return a.row < b.row ;
 33     }
 34 }
 35 
 36 bool cmp1 (black a , black b)
 37 {
 38     if (a.pow < b.pow)
 39         return 1 ;
 40     if (a.pow > b.pow)
 41         return 0 ;
 42     if (a.pow == b.pow ) {
 43         if (a.row == b.row)
 44             return a.col < b.col ;
 45         else
 46             return a.row > b.row ;
 47     }
 48 }
 49 
 50 int main ()
 51 {
 52   //  freopen ("a.txt" , "r" , stdin) ;
 53     for (int i = 1 ; i <= 17 ; i++) {
 54         if (i & 1)
 55             gets (st) ;
 56         else
 57             gets (chess[i / 2]) ;
 58     }
 59     int k , f;
 60 
 61     for (int i = 1 ; i <= 8 ; i++) {
 62         k = 1 ;
 63         for (int j = 2 ; j < 33 ; j += 4 , k++) {
 64             switch ( chess[i][j] )
 65             {
 66                 case 'k' : a[i][k] = 1 ; break ;
 67                 case 'q' : a[i][k] = 2 ; break ;
 68                 case 'r' : a[i][k] = 3 ; break ;
 69                 case 'b' : a[i][k] = 4 ; break ;
 70                 case 'n' : a[i][k] = 5 ; break ;
 71                 case 'p' : a[i][k] = 6 ; break ;
 72                 case 'K' : a[i][k] = -1 ; break ;
 73                 case 'Q' : a[i][k] = -2 ; break ;
 74                 case 'R' : a[i][k] = -3 ; break ;
 75                 case 'B' : a[i][k] = -4 ; break ;
 76                 case 'N' : a[i][k] = -5 ; break ;
 77                 case 'P' : a[i][k] = -6 ; break ;
 78                 case ':' : a[i][k] = 0 ; break ;
 79                 case '.' : a[i][k] = 0 ; break ;
 80                 default : break ;
 81             }
 82         }
 83     }
 84    /* for (int i = 1 ; i <= 8 ; i++) {
 85         for (int j = 1 ; j <= 8 ; j++) {
 86             printf ("%d	" , a[i][j]) ;
 87         }
 88         puts ("") ;
 89     }*/
 90     k = 0 , f = 0 ;
 91     for (int i = 1 ; i <= 8 ; i++) {
 92         for (int j = 1 ; j <= 8 ; j++) {
 93             if (a[i][j] > 0) {
 94                 bl[k].pow = a[i][j] ;
 95                 bl[k].row = '0' + 9 - i ;
 96                 bl[k].col = 'a' + j - 1 ;
 97                 switch (a[i][j])
 98                 {
 99                     case 1 : bl[k].ch = 'K' ; break ;
100                     case 2 : bl[k].ch = 'Q' ; break ;
101                     case 3 : bl[k].ch = 'R' ; break ;
102                     case 4 : bl[k].ch = 'B' ; break ;
103                     case 5 : bl[k].ch = 'N' ; break ;
104                     default : break ;
105                 }
106                 k++ ;
107             }
108             if (a[i][j] < 0) {
109                 wh[f].pow = -a[i][j] ;
110                 wh[f].row = '0' + 9 - i ;
111                 wh[f].col = 'a' + j - 1 ;
112                  switch (-a[i][j])
113                 {
114                     case 1 : wh[f].ch = 'K' ; break ;
115                     case 2 : wh[f].ch = 'Q' ; break ;
116                     case 3 : wh[f].ch = 'R' ; break ;
117                     case 4 : wh[f].ch = 'B' ; break ;
118                     case 5 : wh[f].ch = 'N' ; break ;
119                     default : break ;
120                 }
121                 f++ ;
122             }
123         }
124     }
125 
126     sort (bl , bl + k , cmp1) ;
127     sort (wh , wh + f , cmp2) ;
128 
129     printf ("White: ") ;
130     for (int i = 0 ; i < f ; i++) {
131        // printf ("wh[%d].pow = %d
" , i , wh[i].pow) ;
132         if (wh[i].pow != 6)
133             printf ("%c" , wh[i].ch) ;
134         printf ("%c%c" , wh[i].col , wh[i].row ) ;
135         if (i != f - 1)
136             printf (",") ;
137     }
138     puts ("") ;
139 
140     printf ("Black: ") ;
141     for (int i = 0 ; i < k ; i++) {
142         if (bl[i].pow != 6)
143             printf ("%c" , bl[i].ch) ;
144         printf ("%c%c" , bl[i].col , bl[i].row ) ;
145         if (i != k - 1)
146             printf (",") ;
147     }
148     puts ("") ;
149     return 0 ;
150 }
View Code
原文地址:https://www.cnblogs.com/get-an-AC-everyday/p/4306717.html