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2 years ago
#include <bits/stdc++.h>
using namespace std;
int a[11][11] =
{
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
{1, 0, 0, 1, 0, 0, 0, 1, 0, 1},
{1, 0, 0, 1, 0, 0, 0, 1, 0, 1},
{1, 0, 0, 0, 0, 1, 1, 0, 0, 1},
{1, 0, 1, 1, 1, 0, 0, 0, 0, 1},
{1, 0, 0, 0, 1, 0, 0, 0, 0, 1},
{1, 0, 1, 0, 0, 0, 1, 0, 0, 1},
{1, 0, 1, 1, 1, 0, 1, 1, 0, 1},
{1, 1, 0, 0, 0, 0, 0, 0, 0, 1},
{1, 1, 1, 1, 1, 1, 1, 1, 1, 1}
};
int dir[4][2] = {{-1, 0},
{0, 1},
{1, 0},
{0, -1}};
int vis[11][11];//用来标记有没有走过(有没有在队列中)
int b[11][11];//用来记录bfs的过程
struct Node {
int x, y;
};
queue<Node> q;
int main() {
memset(b, 0, sizeof(b));
memset(vis, 0, sizeof(vis));
while (!q.empty()) q.pop();//初始化队列
Node start;
start.x = 1;
start.y = 1;
vis[1][1] = 1;
b[1][1] = 1;
q.push(start);//把起点放进队列
bool f = 0;
while (!q.empty()) {
Node temp;
temp = q.front();//取出队头元素
q.pop();
int x = temp.x;
int y = temp.y;
if (x == 8 && y == 8)//若走到了
{
f = 1;
Node road[111];//用来记录路径
//for(int i=1;i<=8;i++)//可以通过输出b数组来观察bfs的实现过程
//{
// for(int j=1;j<=8;j++)
// {
// printf("%5d",b[i][j]);
// }
// cout<<endl;
//}
//cout<<b[8][8]<<endl;
int k = 1;
while (!(x == 1 && y == 1))//通过b数组来找到之前是哪一个点走到xy的
{
road[k].x = x;
road[k++].y = y;
for (int i = 0; i < 4; i++) {
int xx = x + dir[i][0];
int yy = y + dir[i][1];
if (xx < 1 || yy < 1 || xx > 8 || yy > 8) continue;//超出范围的不要
if (b[xx][yy] == b[x][y] - 1) {
x = xx;//倒退回去
y = yy;
break;//一定要跳出要把更新的xy放到road里
}
}
}
road[k].x = 1;//别忘了把起点放进去
road[k].y = 1;
for (int i = k; i >= 1; i--)//输出路径
{
cout << road[i].x << " " << road[i].y << endl;
}
}
if (f == 1) //找到路了就不用再跑大循环了
break;
for (int i = 0; i < 4; i++)//遍历四个方向
{
int xx = x + dir[i][0];
int yy = y + dir[i][1];
if (a[xx][yy] == 0 && vis[xx][yy] == 0) {
vis[xx][yy] = 1;
Node New;
New.x = xx;
New.y = yy;
b[xx][yy] = b[x][y] + 1;//用来标记走到xxyy是第几步
q.push(New);//放进队列
}
}
}
return 0;
}