hiho_1054_滑动解锁

题目大意

    智能手机九点屏幕滑动解锁，如果给出某些连接线段，求出经过所有给出线段的合法的滑动解锁手势的总数。题目链接： 
滑动解锁

题目分析

    首先，尝试求解没有给定线段情况下，所有合法的路径的总数。可以使用dfs进行搜索。代码如下：

void dfs(int row, int col, int cur_len) {
	visited[row][col] = true;
	if (cur_len >= 4) { //到达该点时，走过的路径长度大于等于4，则为合法的一个解锁手势
		total_count++;
	}
	if (cur_len == 10)
		return;
	for (int i = 0; i < 3; i++) {
		for (int j = 0; j < 3; j++) {
			int next_row = (row + i) % 3;
			int next_col = (col + j) % 3;
			if (!visited[next_row][next_col]) {
		    //可能出现当前点和下一个点的连线上经过了另一个点的情况，进行判断。如果
		    //经过的另一个点之前被访问过，则这次仍然是合法的。
				if (abs(row - next_row) == 2 && abs(col - next_col) != 1 ||
					abs(col - next_col) == 2 && abs(row - next_row) != 1) {
					int mid_row = (row + next_row) / 2;
					int mid_col = (col + next_col) / 2;
					if (!visited[mid_row][mid_col]) {
						continue;
					}
				}
				int cur = 3 * row + col;
				int next = 3 * next_row + next_col;
					dfs(next_row, next_col, cur_len + 1, need_use, cur_used + 1);
			}
		}
	}
	visited[row][col] = false;
}

在上面的dfs搜索基础上，添加对已有线段的限制。9个点，维护 connected[9][9]， connected[i][j] 表示已经有线段将i和j连接。搜索的时候，还需要维护状态，cur_used表示当前已经经过了已有线段的数目，如果已经经过了所有的线段。且当前路径经过点数大于等于4，则是一个合法的解锁路径。

实现

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
bool visited[3][3];
bool connected[9][9];
int total_count;
//cur_len 为到达row，col点时候，路径的长度； cur_used表示已经走过的路径所覆盖的已有线段的个数
void dfs(int row, int col, int cur_len, int need_use, int cur_used) {
	visited[row][col] = true;
	if (cur_len >= 4 && need_use == cur_used) {
		total_count++;
	}
	if (cur_len == 10)
		return;
	for (int i = 0; i < 3; i++) {
		for (int j = 0; j < 3; j++) {
			int next_row = (row + i) % 3;
			int next_col = (col + j) % 3;
			if (!visited[next_row][next_col]) {
				if (abs(row - next_row) == 2 && abs(col - next_col) != 1 ||
					abs(col - next_col) == 2 && abs(row - next_row) != 1) {
					int mid_row = (row + next_row) / 2;
					int mid_col = (col + next_col) / 2;
					if (!visited[mid_row][mid_col]) {
						continue;
					}
				}
				int cur = 3 * row + col;
				int next = 3 * next_row + next_col;
				if(connected[cur][next])
					dfs(next_row, next_col, cur_len + 1, need_use, cur_used + 1);
				else
					dfs(next_row, next_col, cur_len + 1, need_use, cur_used);
			}
		}
	}
	visited[row][col] = false;
}
void GetCount(int need_use) {
	for (int row = 0; row < 3; row++) {
		for (int col = 0; col < 3; col++) {
			dfs(row, col, 1, need_use, 0);
		}
	}
}
int main() {
	memset(visited, false, sizeof(visited));
	int T, need_use, u, v;
	scanf("%d", &T);
	while (T--) {
		total_count = 0;
		memset(connected, false, sizeof(connected));
		scanf("%d", &need_use);
		for (int i = 0; i < need_use; i++) {
			scanf("%d %d", &u, &v);
			connected[u-1][v-1] = connected[v-1][u-1] = true;
		}
		GetCount(need_use);
		printf("%d
", total_count);
	}	
	return 0;
}