ZOJ 4020 Traffic Light

Posted 2020-10-30 灬从此以后灬

Traffic Light

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Time Limit: 1 Second      Memory Limit: 131072 KB

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DreamGrid City is a city with  intersections arranged into a grid of  rows and  columns. The intersection on the -th row and the -th column can be described as , and two intersections  and  are connected by a road if .

At each intersection stands a traffic light. A traffic light can only be in one of the two states: 0 and 1. If the traffic light at the intersection  is in state 0, one can only move from  to  or ; If the traffic light is in state 1, one can only move from  to  or  (of course, the destination must be another intersection in the city).

BaoBao lives at the intersection , and he wants to visit his best friend DreamGrid living at the intersection . After his departure, in each minute the following things will happen in order:

• BaoBao moves from his current intersection to another neighboring intersection along a road. As a law-abiding citizen, BaoBao has to obey the traffic light rules when moving.
• Every traffic light changes its state. If a traffic light is in state 0, it will switch to state 1; If a traffic light is in state 1, it will switch to state 0.

As an energetic young man, BaoBao doesn‘t want to wait for the traffic lights, and he must move in each minute until he arrives at DreamGrid‘s house. Please tell BaoBao the shortest possible time he can move from  to  to meet his friend, or tell him that this is impossible.

Input

There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:

The first line contains two integers  and  (), indicating the size of the city.

For the following  lines, the -th line contains  integers  (), where  indicates the initial state of the traffic light at intersection .

The next line contains four integers , ,  and  (, ), indicating the starting intersection and the destination intersection.

It‘s guaranteed that the sum of  over all test cases will not exceed .

Output

For each test case output one line containing one integer, indicating the shortest possible time (in minute) BaoBao can move from  to  without stopping. If it is impossible for BaoBao to arrive at DreamGrid‘s house, print "-1" (without quotes) instead.

Sample Input

4
2 3
1 1 0
0 1 0
1 3 2 1
2 3
1 0 0
1 1 0
1 3 1 2
2 2
1 0
1 0
1 1 2 2
1 2
0 1
1 1 1 1

Sample Output

3
5
-1
0

Hint

For the first sample test case, BaoBao can follow this path: .

For the second sample test case, due to the traffic light rules, BaoBao can‘t go from  to  directly. Instead, he should follow this path: .

For the third sample test case, it‘s easy to discover that BaoBao can only go back and forth between  and .

题意  n*m的方格 每个点都有一个状态 状态0 和 1 每秒每个格子都会变成另一个状态   现给出起点 终点 状态为0的点只能上下走 状态为1 的点只能左右走 问起点到终点的最短路长

解析   直接bfs搜就好了 记录当前点的层次 从而判断出现在的状态 再将相应的点加入队列。

　　　因为 n*m<=1e5  所以外边开vector数组 输完n,m之后再开其他数组 

AC代码

 1 #include <stdio.h>
 2 #include <math.h>
 3 #include <string.h>
 4 #include <stdlib.h>
 5 #include <iostream>
 6 #include <sstream>
 7 #include <algorithm>
 8 #include <string>
 9 #include <queue>
10 #include <map>
11 #include <vector>
12 #include <set>
13 #include <utility>
14 #include <stack>
15 using namespace std;
16 const int maxn = 3e5+50,inf = 0x3f3f3f3f, mod = 998244353;
17 const double epx = 1e-6;
18 const double PI = acos(-1.0);
19 typedef long long ll;
20 typedef pair<int,int> pii;
21 const ll INF = 1e18;
22 vector<int> g[maxn];
23 int dir[10][10]= {{1,0},{-1,0},{0,1},{0,-1}};
24 int n,m;
25 int main()
26 {
27     int t;
28     cin>>t;
29     while(t--)
30     {
31         cin>>n>>m;
32         for(int i=0; i<n; i++)
33         {
34             g[i].clear();
35             for(int j=0; j<m; j++)
36             {
37                 int x;
38                 scanf("%d",&x);
39                 g[i].push_back(x);
40             }
41         }
42         int vis[n+10][m+10];
43         int ans[n+10][m+10];
44         memset(vis,0,sizeof(vis));
45         memset(ans,-1,sizeof(ans));
46         int sx,sy,ex,ey;
47         cin>>sx>>sy>>ex>>ey;
48         vis[sx-1][sy-1]=1;
49         ans[sx-1][sy-1]=0;
50         queue<pii> q;
51         q.push(make_pair(sx-1,sy-1));
52         while(!q.empty())
53         {
54             pii temp=q.front();
55             q.pop();
56             int flag;
57             int x=temp.first,y=temp.second;
58             if(ans[x][y]%2==0)
59                 flag=g[x][y];
60             else
61                 flag=!g[x][y];
62             //cout<<ans[x][y]<<" "<<x<<" "<<y<<" "<<flag<<endl;
63             if(flag==0)
64                 for(int i=0; i<2; i++)
65                 {
66                     int x1=x+dir[i][0];
67                     int y1=y+dir[i][1];
68                     if(x1>=0&&x1<n&&y1>=0&&y1<m&&vis[x1][y1]==0)
69                     {
70                         ans[x1][y1]=ans[x][y]+1;
71                         vis[x1][y1]=1;
72                         q.push(make_pair(x1,y1));
73                     }
74                 }
75             else if(flag==1)
76                 for(int i=2; i<4; i++)
77                 {
78                     int x1=x+dir[i][0];
79                     int y1=y+dir[i][1];
80                     if(x1>=0&&x1<n&&y1>=0&&y1<m&&vis[x1][y1]==0)
81                     {
82                         ans[x1][y1]=ans[x][y]+1;
83                         vis[x1][y1]=1;
84                         q.push(make_pair(x1,y1));
85                     }
86                 }
87         }
88         printf("%d\n",ans[ex-1][ey-1]);
89     }
90 }

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