UVA 387 A Puzzling Problem（dfs + 模拟）

A Puzzling Problem

The goal of this problem is to write a program which will take from 1 to 5 puzzle pieces such as those shown below and arrange them, if possible, to form a square. An example set of pieces is shown here.

The pieces cannot be rotated or flipped from their original orientation in an attempt to form a square from the set. All of the pieces must be used to form the square. There may be more than one possible solution for a set of pieces, and not every arrangement will work even with a set for which a solution can be found. Examples using the above set of pieces are shown here.

Input

The input file for this program contains several puzzles (i.e. sets of puzzle pieces) to be solved. The first line of the file is the number of pieces in the first puzzle. Each piece is then specified by listing a single line with two integers, the number of rows and columns in the piece, followed by one or more lines which specify the shape of the piece. The shape specification consists of `0' and `1' characters, with the `1' characters indicating the solid shape of the puzzle (the `0' characters are merely placeholders). For example, piece `A' above would be specified as follows:

2 3
111
101

The pieces should be numbered by the order they are encountered in the puzzle. That is, the first piece in a puzzle is piece #1, the next is piece #2, etc. All pieces may be assumed to be valid and no larger than 4 rows by 4 columns.

The line following the final line of the last piece contains the number of pieces in the next puzzle, again followed by the puzzle pieces and so on. The end of the input file is indicated by a zero in place of the number of puzzle pieces.

Output

Your program should report a solution, if one is possible, in the format shown by the examples below. A 4-row by 4-column square should be created, with each piece occupying its location in the solution. The solid portions of piece #1 should be replaced with `1' characters, of piece #2 with `2' characters, etc. The solutions for each puzzle should be separated by a single blank line.

If there are multiple solutions, any of them is acceptable. For puzzles which have no possible solution simply report ``No solution possible''.

Sample Input

4
2 3
111
101
4 2
01
01
11
01
2 1
1
1
3 2
10
10
11
4
1 4
1111
1 4
1111
1 4
1111
2 3
111
001
5
2 2
11
11
2 3
111
100
3 2
11
01
01
1 3
111
1 1
1
0

Sample Output

1112
1412
3422
3442

No solution possible

1133
1153
2223
2444

题意：输入一个t表示有t块拼图。以下t行输入n，m表示n*m的拼图。然后以下n*m行输入该拼图的形状。要判断这些拼图能不能拼出一个4 * 4的正方形。如果行输出拼的方法（一种即可）如果不行输出No solution possible。注意每块拼图都要用上哦亲。。

思路：这题数据量不是很大。4 * 4。我是直接深搜没有剪枝就过了。。大概就是模拟一个个拼图去放置。放置到一种可以情况就可以结束了。

#include <stdio.h>
#include <string.h>

int t;
int n, m;
char map[10][10];
int judge;
int vis[20];
struct Block {
	char set[10][10];
	int num;
	int n, m;
} b[20];

int ju(int ii, int x, int y) {
	int i, j;
	for (i = x; i < x + b[ii].n; i ++)
		for (j = y; j < y + b[ii].m; j ++)
			if (map[i][j] != '0' && b[ii].set[i -x][j - y] != '0')
				return 0;
	return 1;
}
void dfs(int x, int y, int tt, int num) {
	int i, j, k;
	if (tt == t) {
		if (num == 16)
			judge = 1;
		return;
	}
	if (y > 4) {
		if (x < 4)
			dfs(x + 1, 1, tt, num);
		return;
	}
	for (i = 1; i <= t; i ++) {
		if (!vis[i] && ju(i, x, y)) {
			vis[i] = 1;
			for (j = x; j < x + b[i].n; j ++)
				for (k = y; k < y + b[i].m; k ++)
					if (b[i].set[j - x][k - y] != '0')
						map[j][k] = b[i].set[j - x][k - y] + i - 1;
			dfs(x, y + 1, tt + 1, num + b[i].num);
			if (judge) return;
			vis[i] = 0;
			for (j = x; j < x + b[i].n; j ++)
				for (k = y; k < y + b[i].m; k ++)
					if (b[i].set[j - x][k - y] != '0')
						map[j][k] = '0';
		}
	}
	dfs(x, y + 1, tt, num);
}
int main() {
	int i, j, k;
	int bo = 0;
	while (scanf("%d", &t) != EOF && t) {
		if (bo == 0) 
			bo = 1;
		else
			printf("\n");
		judge = 0;
		memset(b, 0, sizeof(b));
		memset(vis, 0, sizeof(vis));
		memset(map, '1', sizeof(map));
		for (i = 1; i <= 4; i ++)
			for (j = 1; j <= 4; j ++)
				map[i][j] = '0';
		for (i = 1; i <= t ; i ++) {
			scanf("%d%d%*c", &n, &m);
			b[i].n = n; b[i].m = m;
			for (j = 0; j < n; j ++) {
				gets(b[i].set[j]);
				int len = strlen(b[i].set[j]);
				for (k = 0; k < len; k ++)
					if (b[i].set[j][k] != '0')
						b[i].num ++;
			}
		}
		dfs(1, 1, 0, 0);
		if (judge) {
			for (i = 1; i <= 4; i ++) {
				for (j = 1; j <= 4; j ++)
					printf("%c", map[i][j]);
				printf("\n");
			}
		}
		else
			printf("No solution possible\n");
	}
	return 0;
}