這道題的處理方法是把待處理的字符數組邊界設為空格作為出口的標志，另設一個記錄所走步數的數組steps，但是這個題目分兩個走法，其一沒有重復的路徑，遇到空格后直接輸出的步數，用mark = 0標記； 另一種情況就是出現了重復的路徑也就是題目中if ( steps[curi][curj] != 0 )即下一步是已經走了的路徑，那么這必然是循環的開始，根據記錄的step和下一個step值之間的關系就可以輸出
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
int main ()
{
char map[13][13];
int steps[13][13];
int num, col, start;
while ( scanf ("%d %d %d", &num, &col, &start ) != EOF && ( num != 0 && col != 0 && start != 0 ) )
{
getchar ();
memset (map, 0, sizeof(map) );
memset (steps, 0, sizeof(steps) );
for ( int i = 0; i < col + 2; i ++ )
{
map[0][i] = ' ';
map[num + 1][i] = ' ';
}
for ( int i = 0; i < num + 2; i ++ )
{
map[i][0] = ' ';
map[i][col + 1] = ' ';
}
for ( int i = 1; i <= num; i ++)
{
for ( int j = 1; j <= col; j ++ )
{
scanf ("%c", &map[i][j]);
}
getchar ();
}
int step = 1;
int curi = 1; int curj = start;
int mark = 0;
int temp = 1; int t = 0;
while ( map[curi][curj] != ' ' )
{
if ( steps[curi][curj] != 0 )
{
temp = step;
t = steps[curi][curj];
mark = 1;
break;
}
else
{
steps[curi][curj] = step;
if ( map[curi][curj] == 'N' )
{
curi = curi - 1;
curj = curj;
}
else if ( map[curi][curj] == 'S' )
{
curi = curi + 1;
curj = curj;
}
else if ( map[curi][curj] == 'W' )
{
curi = curi;
curj = curj - 1;
}
else if ( map[curi][curj] == 'E' )
{
curi = curi;
curj = curj + 1;
}
step ++;
}
}
if ( mark == 0 )
printf ("%d step(s) to exit\n", step - 1 );
else
printf ("%d step(s) before a loop of %d step(s)\n", t - 1, temp - t );
}
//system ("pause");
return 0;
}
posted on 2010-08-30 15:28
雪黛依夢 閱讀(372)
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