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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
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。
#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\n", total_count);
}
return 0;
}
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原文地址：http://www.cnblogs.com/gtarcoder/p/5538099.html
