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227 Puzzle

2017-01-20 22:22 260 查看

这个题是给出一个5*5的方格里面有一个空格，要求输入各种字符来代表空格的移动并输出移动后的5*5图形，
***这个题给我的收获是如何跳出一个多重循环，比如for(x=0;x<5;x++)for(y=0;y<5;y++)当在25个数据中找到所求的位置的时候，就可以令x和y等于5，这样的话就自动跳出循环了
代码如下

#include <stdio.h>
#include <string.h>
int main () {
int cases = 0;
bool line = false;
char initial [5] [7];
while ( gets (initial [0]) ) {
if ( strcmp (initial [0], "Z") == 0 )
return 0;
gets (initial [1]);
gets (initial [2]);
gets (initial [3]);
gets (initial [4]);
int blank_x;
int blank_y;
for ( int i = 0; i < 5; i++ ) {
for ( int j = 0; j < 5; j++ ) {
if ( initial [i] [j] == ' ' ) {
blank_x = i;
blank_y = j;
i = j = 5;
}
}
}
char command [1000];
bool valid = true;
bool exit_koro = false;
while ( !exit_koro && gets (command)) {
for ( int i = 0; command [i] != 0; i++ ) {
if ( command [i] == '0' || !valid ) {
exit_koro = true;
break;
}
switch (command [i]) {
case 'A' :
if ( blank_x == 0 )
valid = false;
else {
initial [blank_x] [blank_y] = initial [blank_x - 1] [blank_y];
initial [blank_x - 1] [blank_y] = ' ';
blank_x--;
}
break;
case 'B' :
if ( blank_x == 4 )
valid = false;
else {
initial [blank_x] [blank_y] = initial [blank_x + 1] [blank_y];
initial [blank_x + 1] [blank_y] = ' ';
blank_x++;
}
break;
case 'R' :
if ( blank_y == 4 )
valid = false;
else {
initial [blank_x] [blank_y] = initial [blank_x] [blank_y + 1];
initial [blank_x] [blank_y + 1] = ' ';
blank_y++;
}
break;
case 'L' :
if ( blank_y == 0 )
valid = false;
else {
initial [blank_x] [blank_y] = initial [blank_x] [blank_y - 1];
initial [blank_x] [blank_y - 1] = ' ';
blank_y--;
}
break;
}
}
}
if ( line )
printf ("\n");
line = true;
printf ("Puzzle #%d:\n", ++cases);
if ( valid ) {
for ( int i = 0; i < 5; i++ ) {
printf ("%c %c %c %c %c\n", initial [i] [0], initial [i] [1],
initial [i] [2], initial [i] [3], initial [i] [4]);
}
}
else
printf ("This puzzle has no final configuration.\n");
}
return 0;
}

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标签： C语言 oj题移动图形

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