专题一搜索 C - Computer Game

1. 题目
   

   Monocarp is playing a computer game. Now he wants to complete the first level of this game.

   A level is a rectangular grid of 22 rows and nn columns. Monocarp controls a character, which starts in cell (1, 1) — at the intersection of the 11-st row and the 11-st column.

   Monocarp's character can move from one cell to another in one step if the cells are adjacent by side and/or corner. Formally, it is possible to move from cell (x1​,y1​) to cell (x2​,y2​) in one step if ∣x1​−x2​∣≤1 and ∣y1​−y2​∣≤1. Obviously, it is prohibited to go outside the grid.

   There are traps in some cells. If Monocarp's character finds himself in such a cell, he dies, and the game ends.

   To complete a level, Monocarp's character should reach cell (2, n) — at the intersection of row 22 and column nn.

   Help Monocarp determine if it is possible to complete the level.

   Input

   The first line contains a single integer tt (1 \le t \le 1001≤t≤100) — the number of test cases. Then the test cases follow. Each test case consists of three lines.

   The first line contains a single integer nn (3 \le n \le 1003≤n≤100) — the number of columns.

   The next two lines describe the level. The ii-th of these lines describes the ii-th line of the level — the line consists of the characters '0' and '1'. The character '0' corresponds to a safe cell, the character '1' corresponds to a trap cell.

   Additional constraint on the input: cells (1, 1) and (2, n) are safe.

   Output

   For each test case, output YES if it is possible to complete the level, and NO otherwise.

   Example
   Input

   4
   3
   000
   000
   4
   0011
   1100
   4
   0111
   1110
   6
   010101
   101010
   

   Output

   YES
   YES
   NO
   YES
   

   Note

   Consider the example from the statement.

   In the first test case, one of the possible paths is(1,1)→(2,2)→(2,3).

   In the second test case, one of the possible paths is (1,1)→(1,2)→(2,3)→(2,4).

   In the fourth test case, one of the possible paths is (1,1)→(2,2)→(1,3)→(2,4)→(1,5)→(2,6).

2. 思路
   比B题那个跳马样板题还简单，设置5个方向搜就完了
   此外还有个简便方法：某一列两个都是1那么必定不能跳过去，除此之外都行
3. 代码
   

   #include<cstdio>
   #include<cstring>
   #include<cmath>
   #include<algorithm>
   #include<queue>
   using namespace std;
   
   int t,n,nx,ny;
   int vis[3][105];
   int map[3][105];
   int main()
   {
   	scanf("%d",&t);
   	for(int i=0;i<t;i++)
   	{
   		int flag=0;
   		scanf("%d",&n);
   		for(int i=1;i<=2;i++)
   		{
   			for(int j=1;j<=n;j++)
   			{
   				scanf("%1d",&map[i][j]);
   			}
   		}
   		queue<int>qx,qy;
   		qx.push(1),qy.push(1);
   		vis[1][1]=1;
   		while(!qx.empty()&&!qy.empty())
   		{
   			nx=qx.front(),qx.pop();
   			ny=qy.front(),qy.pop();
   			if(nx==2&&ny==n)
   			{
   				flag=1;
   				break;
   			}
   			if(nx-1>=1&&ny+1<=n&&vis[nx-1][ny+1]==0&&map[nx-1][ny+1]==0)
   			{
   				qx.push(nx-1);
   				qy.push(ny+1);
   				vis[nx-1][ny+1]=1;
   			}
   			if(ny+1>=0&&vis[nx][ny+1]==0&&map[nx][ny+1]==0)
   			{
   				qx.push(nx);
   				qy.push(ny+1);
   				vis[nx][ny+1]=1;
   			}
   			if(nx+1<=2&&ny+1<=n&&vis[nx+1][ny+1]==0&&map[nx+1][ny+1]==0)
   			{
   				qx.push(nx+1);
   				qy.push(ny+1);
   				vis[nx+1][ny+1]=1;
   			}
   			if(nx-1>=1&&vis[nx-1][ny]==0&&map[nx-1][ny]==0)
   			{
   				qx.push(nx-1);
   				qy.push(ny);
   				vis[nx-1][ny]=1;
   			}
   			if(nx+1<=2&&vis[nx+1][ny]==0&&map[nx+1][ny]==0)
   			{
   				qx.push(nx+1);
   				qy.push(ny);
   				vis[nx+1][ny]=1;
   			}
   		}
   		memset(vis,0,sizeof(vis));
   		if(flag)
   		{
   			printf("YES\n");
   		}
   		else
   		{
   			printf("NO\n");
   		}
   		memset(map,0,sizeof(map));
   	}
   	return 0;
   }