poj2965解题报告

这道题目的思路和 poj1753的思路一样。
有所不同的是这道题目中还需要输出搜索的路径，于是在unit中加了一个pre变量以记录搜索的路径，最后通过递归调用print_detail从前往后输出宽搜的结果。

Problem: 2965		User: godfrey90
Memory: 1992K		Time: 1000MS
Language: G++		Result: Accepted
Source Code
#include<cstdio>

struct unit {
	int x;
	int rounds;
	int i;
	int pre;
};

void flip(unit a, int n, unit& b);
void print_detail(unit queue[], int n);

int main() {
	/*queue*/
	unit queue[100000];
	queue[0].x = 0;
	queue[0].i = -1;
	queue[0].rounds = 0;
	queue[0].pre=-1;
	int p = 0, q = 0;
	//judge used
	bool used[100000] = { false };
	/*read in*/
	char str[10];
	for (int i = 0; i < 4; i++) {
		scanf("%s", str);
		for (int j = 0; j < 4; j++) {
			if (str[j] == '+') {
				queue[0].x = ((0X1 << (i * 4 + j)) | (queue[0].x));
			}
		}
	}
	/*search*/

	while (!(queue[q].x == 0)) {
		for (int i = 0; i < 16; i++) {
			if (queue[p].i == i)
				continue;
			q++;
			flip(queue[p], i, queue[q]);
			queue[q].pre = p;
			if (used[queue[q].x])
				q--;
			used[queue[q].x] = true;
			if (queue[q].x == 0)
				break;
		}

		if (p == q) {
			printf("Impossible");
			break;
		}

		p++;
	}
	if (queue[q].x == 0)
		printf("%d\n", queue[q].rounds);
	print_detail(queue,q);
	return 0;
}
void flip(unit a, int n, unit& b) {
	int x = n / 4, y = n % 4;
	b.x = a.x;
	b.x = ((b.x) ^ (0X1 << (n)));

	b.x = ((b.x) ^ (0X1 << (y)));
	b.x = ((b.x) ^ (0X1 << (y+4)));
	b.x = ((b.x) ^ (0X1 << (y+4*2)));
	b.x = ((b.x) ^ (0X1 << (y+4*3)));

	b.x = ((b.x) ^ (0X1 << (4*x)));
	b.x = ((b.x) ^ (0X1 << (4*x+1)));
	b.x = ((b.x) ^ (0X1 << (4*x+2)));
	b.x = ((b.x) ^ (0X1 << (4*x+3)));

	b.rounds = a.rounds + 1;
	b.i = n;
}
void print_detail(unit queue[], int n){
	if(queue[n].pre==-1) return;
	print_detail(queue,queue[n].pre);
	int x,y;
	x=queue[n].i/4+1;
	y=queue[n].i%4+1;
	printf("%d %d\n",x,y);
}