题目

Vasya plays a computer game with ninjas. At this stage Vasya's ninja should get out of a deep canyon.

The canyon consists of two vertical parallel walls, their height is

Initially the ninja is on the lower area of the left wall. He can use each second to perform one of the following actions:

climb one area up;
climb one area down;
jump to the opposite wall. That gets the ninja to the area that is exactly

If at some point of time the ninja tries to get to an area with a number larger than

The canyon gets flooded and each second the water level raises one meter. Initially the water level is at the lower border of the first area. Ninja cannot be on the area covered by water. We can assume that the ninja and the water "move in turns" — first the ninja performs some action, then the water raises for one meter, then the ninja performs one more action and so on.

The level is considered completed if the ninja manages to get out of the canyon.

After several failed attempts Vasya started to doubt whether it is possible to complete the level at all. Help him answer the question.

输入输出格式

输入格式：

The first line contains two integers $1<=n,k<=10^{5}1<=n,k<=105 ) — the height of the canyon and the height of ninja's jump, correspondingly.$

The second line contains the description of the left wall — a string with the length of

The third line describes the right wall in the same format.

It is guaranteed that the first area of the left wall is not dangerous.

输出格式：

Print "YES" (without the quotes) if the ninja can get out from the canyon, otherwise, print "NO" (without the quotes).

输入输出样例

输入样例#1：复制

7 3
---X--X
-X--XX-

输出样例#1：复制

YES

输入样例#2：复制

6 2
--X-X-
X--XX-

输出样例#2：复制

NO

说明

In the first sample the ninja should first jump to the right wall, then go one meter down along the right wall, then jump to the left wall. The next jump can get the ninja from the canyon.

In the second sample there's no way the ninja can get out of the canyon.

DFS模型不必多说，重点是调用dfs时的顺序；

先是顺序

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int N=1e5+200;
 5 
 6 int n,k;
 7 char a[2][N];
 8 bool b[2][N];
 9 
10 bool dfs(bool side,int h,int step){//side代表左右两边墙
11     if (h>n)//jump out of canyon
12         return 1;
13     if (b[side][h]||a[side][h]=='X'||h<step)//h<step 从大峡谷(ノ｀Д)ノ粗去了
14         return 0;
15         
16     b[side][h]=1;//走过了的点
17     return dfs(side^1,h+k,step+1)||dfs(side,h-1,step+1)||dfs(side,h+1,step+1);//顺序很重要，先h+k
18 }
19 int main()
20 {
21 
22     ios_base::sync_with_stdio(0);
23     cin.tie(0);
24     cout.tie(0);//为什么对这三行情有独钟呢？当然是~~~（省略）
25     cin>>n>>k;
26 
27     scanf("%s %s",a[0]+1,a[1]+1);//注意这个输入方法；用于输入两行二维字符数组很合适，
28     //或者：
29     //for(int i=0;i<2;i++){
30     //  for(int j=1;j<=n;j++){
31     //      a[i][j]=getchar()=='X';
32     //}
33     //for(;getchar()^'/n';)
34     //}
35 
36     if(dfs(0,1,1))//
37         cout<<"YES";
38     else
39         cout<<"NO";
40 
41     return 0;
42 }

我们的宗旨是！dfs+思维~~

CF198B Jumping on Walls （DFS+剪枝）

题目

输入输出格式

输入输出样例

说明

猜你喜欢