题目描述
有一个n*m格的迷宫(表示有n行、m列)，其中有可走的也有不可走的，如果用1表示可以走，0表示不可以走，文件读入这n*m个数据和起始点、结束点(起始点和结束点都是用两个数据来描述的，分别表示这个点的行号和列号)。现在要你编程找出所有可行的道路，要求所走的路中没有重复的点，走时只能是上下左右四个方向。如果一条路都不可行，则输出相应信息(用-l表示无路)。
请统一用 左上右下的顺序拓展，也就是 (0,-1),(-1,0),(0,1),(1,0)
输入
第一行是两个数n，m( 1 < n ， m < 15 )，接下来是m行n列由1和0组成的数据，最后两行是起始点和结束点。
输出
所有可行的路径，描述一个点时用(x，y)的形式，除开始点外，其他的都要用“->”表示方向。
如果没有一条可行的路则输出-1。
样例输入
5 6 1 0 0 1 0 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 0 1 1 1 1 5 6
样例输出
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(2,5)->(3,5)->(3,4)->(3,3)->(4,3)->(4,4)->(4,5)->(5,5)->(5,6) (1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(2,5)->(3,5)->(3,4)->(4,4)->(4,5)->(5,5)->(5,6) (1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(2,5)->(3,5)->(4,5)->(5,5)->(5,6) (1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(3,3)->(4,3)->(4,4)->(4,5)->(5,5)->(5,6) (1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(3,5)->(4,5)->(5,5)->(5,6) (1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(4,4)->(4,5)->(5,5)->(5,6) (1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,4)->(2,4)->(2,5)->(3,5)->(4,5)->(5,5)->(5,6) (1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,4)->(3,5)->(4,5)->(5,5)->(5,6) (1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,4)->(4,4)->(4,5)->(5,5)->(5,6) (1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(4,3)->(4,4)->(3,4)->(2,4)->(2,5)->(3,5)->(4,5)->(5,5)->(5,6) (1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(4,3)->(4,4)->(3,4)->(3,5)->(4,5)->(5,5)->(5,6) (1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(4,3)->(4,4)->(4,5)->(5,5)->(5,6)
一，我之前错误的方法
总结：定位到结点A，然后在A结点上下左右搜索，搜索到下一个能走的结点B就把A结点加入序列中，是一种退缩式方法，但缺陷是如果回溯时会出现混乱。
错误代码如下：
void DFS(int x,int y) {
if (hashMap[x][y - 1] == 0 && road[x][y - 1] == 1) {
hashMap[x][y-1] = 1;
point[count].x = x;
point[count].y = y-1;
count++;
DFS(x, y - 1);
}
if (hashMap[x-1][y] == 0 && road[x-1][y] == 1) {
hashMap[x-1][y] = 1;
point[count].x = x-1;
point[count].y = y;
count++;
DFS(x-1, y);
}
if (hashMap[x][y + 1] == 0 && road[x][y + 1] == 1) {
hashMap[x][y+1] = 1;
point[count].x = x;
point[count].y = y+1;
count++;
DFS(x, y + 1);
}
if (hashMap[x+1][y] == 0 && road[x+1][y] == 1) {
hashMap[x+1][y] = 1;
point[count].x = x+1;
point[count].y = y;
count++;
DFS(x+1, y);
}
hashMap[x][y] = 0;
count--;
point[count].x = 0;
point[count].y = 0;
}二，正确的方法：
先定位到结点A，在A结点上下左右搜索，搜索到可走结点B，就把B加入到序列中。
即如果未来结点可走，立马把未来结点加入到序列中，在每一个DFS未来结点之后再把这个序列中的未来结点清空，这一点同全排列
请仔细体会以上思想。
#include<cstdio>
using namespace std;
//到15:20
const int max_n = 289;
int count = 0;
int real_count = 0;
int start_x = 0, start_y = 0;
int end_x = 0, end_y = 0;
int road[17][17] = { 0 };
int hashMap[17][17] = { 0 };
struct Point {
int x, y;
}point[max_n];
void check_print(int x,int y) {
if (end_x == x && y == end_y) {
point[count].x = x;
point[count].y = y;
count++;
for (int i = 0; i < count; i++) {
printf("(%d,%d)", point[i].x, point[i].y);
if (i != count - 1) {
printf("->");
}
else {
printf("\n");
}
}
count--;
point[count].x = 0;
point[count].y = 0;
real_count++;
}
}
void DFS(int x,int y) { //二维数组作为形参不能忽略第二位的大小
if (hashMap[x][y - 1] == 0 && road[x][y - 1] == 1) {
if (x == end_x&&y - 1 == end_y) {
check_print(x, y - 1);
return;
}
hashMap[x][y-1] = 1;
point[count].x = x;
point[count].y = y-1;
count++;
DFS(x, y - 1);
hashMap[x][y-1] = 0;
count--;
point[count].x = 0;
point[count].y = 0;
}
if (hashMap[x-1][y] == 0 && road[x-1][y] == 1) {
if (x-1 == end_x&&y == end_y) {
check_print(x-1, y);
return;
}
hashMap[x-1][y] = 1;
point[count].x = x-1;
point[count].y = y;
count++;
DFS(x-1, y);
hashMap[x-1][y] = 0;
count--;
point[count].x = 0;
point[count].y = 0;
}
if (hashMap[x][y + 1] == 0 && road[x][y + 1] == 1) {
if (x == end_x&&y + 1 == end_y) {
check_print(x, y + 1);
return;
}
hashMap[x][y+1] = 1;
point[count].x = x;
point[count].y = y+1;
count++;
DFS(x, y + 1);
hashMap[x][y+1] = 0;
count--;
point[count].x = 0;
point[count].y = 0;
}
if (hashMap[x+1][y] == 0 && road[x+1][y] == 1) {
if (x+1 == end_x&&y == end_y) {
check_print(x+1, y);
return;
}
hashMap[x+1][y] = 1;
point[count].x = x+1;
point[count].y = y;
count++;
DFS(x+1, y);
hashMap[x+1][y] = 0;
count--;
point[count].x = 0;
point[count].y = 0;
}
/* hashMap[x][y] = 0;
count--;
point[count].x = 0;
point[count].y = 0;*/
}
int main() {
int rows = 0, cols = 0;
int check = 0;
scanf("%d %d", &rows, &cols);
for (int i = 1; i <= rows; i++) {
for (int j = 1; j <= cols; j++) {
scanf("%d", &road[i][j]);
}
}
scanf("%d %d", &start_x, &start_y);
scanf("%d %d", &end_x, &end_y);
hashMap[start_x][start_y] = 1;
point[count].x = start_x;
point[count].y = start_y;
count++;
DFS(start_x, start_y);
if (real_count == 0) {
printf("-1");
}
return 0;
}