UVA 10422 - Knights in FEN

题目大意：初始状态唯一（空格处于中间，题目有给出）， 给出一个状态，问在 10 个移动以内可否到达

解题思路：每次确定dfs的深度，将所有可能收索一遍
#include <cstdio>

char board[10][10] = {"11111", "01111", "00 11", "00001", "00000"}, data[10][10];
int step, dir[8][2] = {{2,1}, {2, -1}, {-2, 1}, {-2, -1}, {1, 2}, {1, -2}, {-1, 2}, {-1, -2}};

bool DFS(int cur, int x, int y) {
if (cur == step) {
for (int i = 0; i < 5; i++)
for (int j = 0; j < 5; j++)
if (board[i][j] != data[i][j])
return false;
return true;
}

for (int i = 0; i < 8; i++)
if (x + dir[i][0] < 5 && x + dir[i][0] >= 0 && y + dir[i][1] < 5 && y + dir[i][1] >= 0) {
int a = x + dir[i][0];
int b = y + dir[i][1];
data[x][y] = data[a][b];
data[a][b] = ' ';
if (DFS(cur + 1, a, b))
return true;
data[a][b] = data[x][y];
data[x][y] = ' ';
}

return false;
}

int main() {
int t, line, column;
scanf("%d\n", &t);
while (t--) {
for (int i = 0; i < 5; i++, getchar())
for (int j = 0; j < 5; j++) {
scanf("%c", &data[i][j]);
if (data[i][j] == ' ') {
line = i;
column = j;
}
}

for (step = 0; step < 11; step++)
if (DFS(0, line, column))
break;

if (step < 11)
printf("Solvable in %d move(s).\n", step);
else
printf("Unsolvable in less than 11 move(s).\n");
}

return 0;
}