Vitaly works at the warehouse. The warehouse can be represented as a grid of?n?×?mcells, each of which either is free or is occupied by a container. From every free cell it's possible to reach every other free cell by moving only through the cells sharing a side. Besides that, there are two robots in the warehouse. The robots are located in different free cells.

Vitaly wants to swap the robots. Robots can move only through free cells sharing a side, moreover, they can't be in the same cell at the same time or move through each other. Find out if the swap can be done.

Input

The first line contains two positive integers?n?and?m?(2?≤?n·m?≤?200000)?— the sizes of the warehouse.

Each of the next?n?lines contains?m?characters. The?j-th character of the?i-th line is ?.? if the corresponding cell is free, ?#? if there is a container on it, ?1? if it's occupied by the first robot, and ?2? if it's occupied by the second robot. The characters ?1? and ?2? appear exactly once in these lines.

Output

Output ?YES? (without quotes) if the robots can be swapped, and ?NO? (without quotes) if that can't be done.

Example

Input 5 3

###
#1#
#.#
#2#
### Output NO Input 3 5

#...#
#1.2#
##### Output YES
題意：1和2要進行位置的交換，問你是否能夠成功？
思路：如果在1能夠到達2，并且他們之間的存在有一個點能夠連接三個方向，或者只有一個方向的點的個數不是2個，則是YES，否則是NO。 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <stack> 5 #include <vector> 6 #include <algorithm> 7 #include<queue> 8 using namespace std; 9 const int maxn=200005; 10 string s[maxn]; 11 typedef pair<int,int>P; 12 P p; 13 queue<P>que; 14 vector<int>vis[maxn]; 15 int n,m,sx,sy,ex,ey; 16 int a[4][2]={{0,1},{0,-1},{1,0},{-1,0}}; 17 void bfs() 18 { 19 for(int i=0; i<n; i++) 20 { 21 vis[i].clear(); 22 for(int j=0; j<m; j++) 23 vis[i].push_back(0); 24 } 25 while(!que.empty()) 26 que.pop(); 27 vis[sx][sy]=1; 28 p.first=sx,p.second=sy; 29 que.push(p); 30 int flag=0; 31 while(!que.empty()) 32 { 33 p=que.front(); 34 que.pop(); 35 int tx=p.first; 36 int ty=p.second; 37 for(int i=0; i<4; i++) 38 { 39 int xx=tx+a[i][0],yy=ty+a[i][1]; 40 if(!(0<=xx&&xx<n&&0<=yy&&yy<m)||s[xx][yy]=='#'||vis[xx][yy]) 41 continue; 42 vis[xx][yy]=1; 43 p.first=xx; 44 p.second=yy; 45 que.push(p); 46 } 47 } 48 } 49 int main() 50 { 51 while(~scanf("%d%d",&n,&m)) 52 { 53 for(int i=0; i<n; i++) 54 { 55 cin>>s[i]; 56 for(int j=0; j<m; j++) 57 { 58 if(s[i][j]=='1') 59 { 60 sx=i,sy=j; 61 } 62 else if(s[i][j]=='2') 63 { 64 ex=i,ey=j; 65 } 66 } 67 } 68 bfs(); 69 if(!vis[ex][ey]) 70 { 71 printf("NO\n"); 72 continue; 73 } 74 int flag=0,ans=0,cnt; 75 for(int i=0;i<n;i++) 76 { 77 for(int j=0;j<m;j++) 78 { 79 if(flag) 80 break; 81 cnt=0; 82 if(vis[i][j]) 83 { 84 for(int k=0;k<4;k++) 85 { 86 if(!(0<=i+a[k][0]&&i+a[k][0]<n&&0<=j+a[k][1]&&j+a[k][1]<m)||s[i+a[k][0]][j+a[k][1]]=='#') 87 continue; 88 if(vis[i+a[k][0]][j+a[k][1]]) 89 cnt++; 90 } 91 if(cnt>2) 92 { 93 flag=1; 94 printf("YES\n"); 95 break; 96 } 97 if(cnt==1) ans++; 98 } 99 } 100 if(flag) 101 break; 102 } 103 if(flag==0) 104 { 105 if(ans==2) printf("NO\n"); 106 else printf("YES\n"); 107 } 108 } 109 return 0; 110 } View Code

知識點：

由于n，m<=200000,所以無法定義flag數組進行標記。使用vector容器能夠很好的解決這一問題。

轉載于:https://www.cnblogs.com/wang-ya-wei/p/6894941.html

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