uva 10422 Knights in FEN（迭代dfs）

Knights in FEN

Input: standard input

Output: standard output

Time Limit: 10 seconds

 

There are black and white knights on a 5 by 5 chessboard. There are twelve of each color, and there is one square that is empty. At any time, a knight can move into an empty square as long as it moves like a knight in normal chess (what else did you expect?).

Given an initial position of the board, the question is: what is the minimum number of moves in which we can reach the final position which is:

Input

First line of the input file contains an integer N (N<14) that indicates how many sets of inputs are there. The description of each set is given below:

Each set consists of five lines; each line represents one row of a chessboard. The positions occupied by white knights are marked by 0 and the positions occupied by black knights are marked by 1. The space corresponds to the empty square on board.

There is no blank line between the two sets of input.

The first set of the sample input below corresponds to this configuration:

Output

For each set your task is to find the minimum number of moves leading from the starting input configuration to the final one. If that number is bigger than 10, then output one line stating

Unsolvable in less than 11 move(s).

 

otherwise output one line stating

Solvable in n move(s).

where n <= 10.

The output for each set is produced in a single line as shown in the sample output.

Sample Input

2

01011

110 1

01110

01010

00100

10110

01 11

10111

01001

00000

Sample Output

Unsolvable in less than 11 move(s).

Solvable in 7 move(s).

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7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 <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

const char begin[5][5] = {'1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '0', '0', ' ', '1', '1', '0', '0', '0', '0', '1', '0', '0', '0', '0', '0'};
const int dx[] = {-2, -2, 2, 2, -1, 1, -1, 1};
const int dy[] = {-1, 1, -1, 1, -2, -2, 2, 2};

char r[5][5];
int ans;
bool flag;

int diff() {
    int sum = 0;
    for (int i = 0; i < 5; i++)
	for (int j = 0; j < 5; j++)
	    sum += (r[i][j] != begin[i][j]);
    return sum / 2;
}

void dfs(int deep, int x, int y) {
    if (deep >= ans) {
	if (memcmp(r, begin, 25 * sizeof(char)) == 0)
	    flag = true;
	return;
    }
    if (deep + diff() > ans)
	return ;
    int swap;
    for (int i = 0; i < 8 && !flag; i++) {
	int newX = x + dx[i], newY = y + dy[i];
	if (newX < 0 || newX >= 5 || newY < 0 || newY >= 5)
	    continue;

	swap = r[x][y];
	r[x][y] = r[newX][newY];
	r[newX][newY] = swap; 

	dfs(deep + 1, newX, newY);

	swap = r[x][y];
	r[x][y] = r[newX][newY];
	r[newX][newY] = swap; 
    }
}

int main() {
    int cas;
    scanf("%d%*c", &cas);
    while (cas--) {
	int x, y;
	for (int i = 0; i < 5; i++) {
	    for (int j = 0; j < 5; j++) {
		scanf("%c", &r[i][j]);
		if (r[i][j] == ' ') {
		    x = i;
		    y = j;
		}
	    }
	    getchar();
	}

	flag = 0;
	for (ans = 0; ans <= 10; ans++) {
	    dfs(0, x, y);
	    if (flag)
		break;
	}

	if (flag)
	    printf("Solvable in %d move(s).\n", ans);
	else
	    printf("Unsolvable in less than 11 move(s).\n");
    }
return 0;
}