UVA - 519 Puzzle (II)

题目大意：给出一些拼图的碎片，要求判断是否能组成一个矩形，对于碎片，F表示平面，I表示凸面，O表示凹面，对于矩形，边必须为平， 并且碎片相连只能是以IO形式

解题思路：回溯+剪枝

剪枝1：在DFS前如果凹凸不等或F数量不等周长就直接输出NO

剪枝2：排序，在判断是否放置当前拼图的时候如果不可以放置，和他一样的不用再判断一次

剪枝3：在递归完成后如果已经成功就直接return

#include <cstdio>
#include <cstring>
#include <algorithm>

char pi[36][5], data[6][6][5];
int num, vis[40], row, col;

int cmp(const void*a, const void*b) {
	return strcmp((char*)a, (char*)b);
}

bool judge(int x, int y, int i) {
	if (x == 0 && pi[i][0] != 'F'|| y == 0 && pi[i][3] != 'F' || y == col - 1 && pi[i][1] != 'F' || x != row - 1 && pi[i][2] == 'F')
		return false;
	if ((x != 0 && pi[i][0] + data[x - 1][y][2] != 'I' + 'O') || (y != 0 && pi[i][3] + data[x][y - 1][1] != 'I' + 'O'))
		return false;

	return true;
}

bool DFS(int x, int y, int cur) {
	if (cur == num)
		return true;

	char temp[5] = {0};
	for (int i = 0; i < num; i++)
		if (!vis[i] && strcmp(temp, pi[i]) && judge(x, y, i)) {
			strcpy(temp, pi[i]);
			strcpy(data[x][y], pi[i]);
			vis[i] = 1;
			if (DFS((cur + 1) / col, (cur + 1) % col, cur + 1))
				return true;
			vis[i] = 0;
		}

	return false;
}

int main() {
	while (scanf("%d%d\n", &row, &col), num = row * col) {
		int edge[120] = {0};
		for (int i = 0; i < num; i++) {
			scanf("%s", pi[i]);
			for (int j = 0; j < 4; j++)
				edge[pi[i][j]]++; 
		}
		qsort(pi, num, sizeof(pi[0]), cmp);
		memset(vis, 0, sizeof(vis));

		printf(edge['F'] == 2 * (row + col) && edge['I'] == edge['O'] && DFS(0, 0, 0) ? "YES\n" : "NO\n");
	}
	return 0;
}