Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20352 Accepted Submission(s): 4946
出队入队具体操作{
出队队首元素 && pop();
队首元素 包括{r, c, dir, num};
遍历该点的四个方向{
判断是否越界 && 可行
不越界&&可行 {
if(转弯数超限)
continue;
if(已走过 && 当前转弯数比之前走过的要多)
continue;
if(已经到达终点)
返回true
对应的vis标记为true;
记录走到该点该方向上的转弯数;
记录入队 ;
}
}
}
#include
#include
#include
#include
using namespace std;
int dx[] = {1, 0, -1, 0}; //房前位置的四个方向
int dy[] = {0, 1, 0, -1};
int n, m;
int k, c1, c2, r1, r2;
int dis[101][101][4];
bool vis[101][101][4];
char map[101][101];
struct Node {
int r, c;
int dir, num;
Node(int r, int c, int dir, int num) : r(r), c(c), dir(dir), num(num) {}
};
bool BFS() {
queue Q;
memset(vis, 0, sizeof(vis));
for(int i = 0; i < 4; i++) {
int x = r1 + dx[i], y = c1 + dy[i];
if(x == r2 && y == c2)
return true;
if(x >= 1 && x <= n && y >= 1 && y <= m && map[x][y] == '.') {
Q.push(Node(x, y, i, 0));
vis[x][y][i] = true;
}
}
while(!Q.empty()) {
Node node = Q.front();
Q.pop(); //记得删除队首
int r = node.r, c = node.c, dir = node.dir, num = node.num;
for(int i = 0; i < 4; i++) {
int x = r + dx[i];
int y = c + dy[i];
if(x >= 1 && x <= n && y >= 1 && y <= m && map[x][y] == '.') {
int n_num = i == dir ? num : num + 1; //判断是否转向
if(n_num > k)
continue;
if(vis[x][y][i] && n_num >= dis[x][y][i])
continue;
if(x == r2 && y == c2)
return true;
vis[x][y][i] = true;
dis[x][y][i] = n_num;
Q.push(Node(x, y, i, n_num));
}
}
}
return false;
}
int main() {
int t;
scanf("%d",&t);
while(t--) {
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= m; j++)
scanf(" %c", &map[i][j]);
}
scanf("%d%d%d%d%d", &k, &c1, &r1, &c2, &r2); //注意输入顺序
if(BFS())
printf("yes\n");
else
printf("no\n");
}
return 0;
}