题意简述
将一个棋盘翻译为文字表达。如下：
+---+---+---+---+---+---+---+---+ |.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.| +---+---+---+---+---+---+---+---+
大写字母表示白方，小写表示黑方。输出时按照KQRBNP的顺序;棋盘格子编号的表示：从下往上是1～8,从左往右是a~h。
遇到相同棋子，白方表示时按从下往上从左往右的顺序，黑方按从上往下从左往右的顺序。
输出格式：
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
算法分析
直接模拟。
分行读入棋盘，分析该行字母，将对应坐标记录。此处忽略黑白相同棋子从上往下和从下往上的区别，只需满足从上往下从左往右。
输出时黑子按存储的顺序输出即可，白子输出相同棋子时判断，若有相同棋子则先输出行数大的。
具体实现见代码。
Problem Status: AC。时间0ms，内存388k
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注意棋子是有规律的，最多16个，但可以不全出现。KQ各一个，RBN最多两个，P8个。
程序样例
#include<stdio.h>
void outputb(int a[16][2]) {
int i;
if (a[0][0]) printf("K%c%d", 'a' + a[0][1] - 1, 9 - a[0][0]);
if (a[1][0]) printf(",Q%c%d", 'a' + a[1][1] - 1, 9 - a[1][0]);
if (a[2][0]) printf(",R%c%d", 'a' + a[2][1] - 1, 9 - a[2][0]);
if (a[3][0]) printf(",R%c%d", 'a' + a[3][1] - 1, 9 - a[3][0]);
if (a[4][0]) printf(",B%c%d", 'a' + a[4][1] - 1, 9 - a[4][0]);
if (a[5][0]) printf(",B%c%d", 'a' + a[5][1] - 1, 9 - a[5][0]);
if (a[6][0]) printf(",N%c%d", 'a' + a[6][1] - 1, 9 - a[6][0]);
if (a[7][0]) printf(",N%c%d", 'a' + a[7][1] - 1, 9 - a[7][0]);
for (i = 8; i < 16; i++)
if (a[i][0]) printf(",%c%d", 'a' + a[i][1] - 1, 9 - a[i][0]);
}
void outputw(int a[16][2]) {
int i, j;
if (a[0][0]) printf("K%c%d", 'a' + a[0][1] - 1, 9 - a[0][0]);
if (a[1][0]) printf(",Q%c%d", 'a' + a[1][1] - 1, 9 - a[1][0]);
if (a[2][0] < a[3][0]) {
if (a[3][0]) printf(",R%c%d", 'a' + a[3][1] - 1, 9 - a[3][0]);
if (a[2][0]) printf(",R%c%d", 'a' + a[2][1] - 1, 9 - a[2][0]);
} else {
if (a[2][0]) printf(",R%c%d", 'a' + a[2][1] - 1, 9 - a[2][0]);
if (a[3][0]) printf(",R%c%d", 'a' + a[3][1] - 1, 9 - a[3][0]);
}
if (a[4][0] < a[5][0]) {
if (a[5][0]) printf(",B%c%d", 'a' + a[5][1] - 1, 9 - a[5][0]);
if (a[4][0]) printf(",B%c%d", 'a' + a[4][1] - 1, 9 - a[4][0]);
} else {
if (a[4][0]) printf(",B%c%d", 'a' + a[4][1] - 1, 9 - a[4][0]);
if (a[5][0]) printf(",B%c%d", 'a' + a[5][1] - 1, 9 - a[5][0]);
}
if (a[6][0] < a[7][0]) {
if (a[7][0]) printf(",N%c%d", 'a' + a[7][1] - 1, 9 - a[7][0]);
if (a[6][0]) printf(",N%c%d", 'a' + a[6][1] - 1, 9 - a[6][0]);
} else {
if (a[6][0]) printf(",N%c%d", 'a' + a[6][1] - 1, 9 - a[6][0]);
if (a[7][0]) printf(",N%c%d", 'a' + a[7][1] - 1, 9 - a[7][0]);
}
for (j = 8; j > 0; j--)
for (i = 8; i < 16; i++)
if (a[i][0] && a[i][0] == j) printf(",%c%d", 'a' + a[i][1] - 1, 9 - a[i][0]);
}
int main() {
char t[50];
int i, j, p = 0, p1 = 0;
int white[16][2] = {
0
}, black[16][2] = {
0
};
scanf("%s", t);
for (i = 0; i < 8; i++) {
scanf("%s", t);
for (j = 2; j < 33; j = j + 4) {
switch (t[j]) {
case 'k':
black[0][0] = i + 1;
black[0][1] = (j - 2) / 4 + 1;
break;
case 'q':
black[1][0] = i + 1;
black[1][1] = (j - 2) / 4 + 1;
break;
case 'r':
if (black[2][0]) {
black[3][0] = i + 1;
black[3][1] = (j - 2) / 4 + 1;
break;
} else {
black[2][0] = i + 1;
black[2][1] = (j - 2) / 4 + 1;
break;
}
case 'b':
if (black[4][0]) {
black[5][0] = i + 1;
black[5][1] = (j - 2) / 4 + 1;
break;
} else {
black[4][0] = i + 1;
black[4][1] = (j - 2) / 4 + 1;
break;
}
case 'n':
if (black[6][0]) {
black[7][0] = i + 1;
black[7][1] = (j - 2) / 4 + 1;
break;
} else {
black[6][0] = i + 1;
black[6][1] = (j - 2) / 4 + 1;
break;
}
case 'p':
black[8 + p][0] = i + 1;
black[8 + p][1] = (j - 2) / 4 + 1;
p++;
break;
//white
case 'K':
white[0][0] = i + 1;
white[0][1] = (j - 2) / 4 + 1;
break;
case 'Q':
white[1][0] = i + 1;
white[1][1] = (j - 2) / 4 + 1;
break;
case 'R':
if (white[2][0]) {
white[3][0] = i + 1;
white[3][1] = (j - 2) / 4 + 1;
break;
} else {
white[2][0] = i + 1;
white[2][1] = (j - 2) / 4 + 1;
break;
}
case 'B':
if (white[4][0]) {
white[5][0] = i + 1;
white[5][1] = (j - 2) / 4 + 1;
break;
} else {
white[4][0] = i + 1;
white[4][1] = (j - 2) / 4 + 1;
break;
}
case 'N':
if (white[6][0]) {
white[7][0] = i + 1;
white[7][1] = (j - 2) / 4 + 1;
break;
} else {
white[6][0] = i + 1;
white[6][1] = (j - 2) / 4 + 1;
break;
}
case 'P':
white[8 + p1][0] = i + 1;
white[8 + p1][1] = (j - 2) / 4 + 1;
p1++;
break;
}
}
scanf("%s", t);
}
printf("White: ");
outputw(white);
printf("\nBlack: ");
outputb(black);
printf("\n");
system("pause");
return 0;
}