Robots at Warehouse（搜索+vector的使用）

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
题意：1和2要进行位置的交换，问你是否能够成功？
思路：如果在1能够到达2，并且他们之间的存在有一个点能够连接三个方向，或者只有一个方向的点的个数不是2个，则是YES，否则是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

知识点：

由于n，m<=200000,所以无法定义flag数组进行标记。使用vector容器能够很好的解决这一问题。

posted @ 2017-05-23 17:04  Wally的博客  阅读(183)  评论(0)  编辑  收藏  举报