hdu 3085

Nightmare Ⅱ

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 720    Accepted Submission(s): 136

Problem Description

Last night, little erriyue had a horrible nightmare. He dreamed that he and his girl friend were trapped in a big maze separately. More terribly, there are two ghosts in the maze. They will kill the people. Now little erriyue wants to know if he could find his girl friend before the ghosts find them.
You may suppose that little erriyue and his girl friend can move in 4 directions. In each second, little erriyue can move 3 steps and his girl friend can move 1 step. The ghosts are evil, every second they will divide into several parts to occupy the grids within 2 steps to them until they occupy the whole maze. You can suppose that at every second the ghosts divide firstly then the little erriyue and his girl friend start to move, and if little erriyue or his girl friend arrive at a grid with a ghost, they will die.
Note: the new ghosts also can devide as the original ghost.

 

 

Input

The input starts with an integer T, means the number of test cases.
Each test case starts with a line contains two integers n and m, means the size of the maze. (1The next n lines describe the maze. Each line contains m characters. The characters may be:
‘.’ denotes an empty place, all can walk on.
‘X’ denotes a wall, only people can’t walk on.
‘M’ denotes little erriyue
‘G’ denotes the girl friend.
‘Z’ denotes the ghosts.
It is guaranteed that will contain exactly one letter M, one letter G and two letters Z. 

 

 

Output

Output a single integer S in one line, denotes erriyue and his girlfriend will meet in the minimum time S if they can meet successfully, or output -1 denotes they failed to meet.

 

 

Sample Input

 

3
5 6
XXXXXX
XZ..ZX
XXXXXX
M.G...
......
5 6
XXXXXX
XZZ..X
XXXXXX
M.....
..G...

10 10
..........
..X.......
..M.X...X.
X.........
.X..X.X.X.
.........X
..XX....X.
X....G...X
...ZX.X...
...Z..X..X

Sample Output

1

1

-1

 

 

Author

二日月

 

 

Source

HDU 2nd “Vegetable-Birds Cup” Programming Open Contest

 

注意的地方

1:鬼可以穿过墙。

2:每一步鬼要先走，如果发现此事G,M被覆盖了，说明，呵呵，他（她）走不了。

   You can suppose that at every second the ghosts divide firstly.

   不然第三组测试数据，不好解释。

3.关于M的三步，很显然，如果前一步走不了，就不能在从这一点继续走下去了。

   就是要注意，如果走一步的时候不能走，那么就不能在这个点基础上又走一步。 

  1 #include
  2 #include
  3 #include
  4 #include
  5 #include
  6 using namespace std;
  7 
  8 int n,m,step;
  9 int zx1,zy1,zx2,zy2;
 10 int mx,my,gx,gy;
 11 int to[4][2]={{1,0},{0,1},{0,-1},{-1,0}};
 12 char a[810][810];
 13 bool hash[2][801][801],glag;
 14 struct node
 15 {
 16     int x,y;
 17 };
 18 queueQ[2];
 19 
 20 int Min(int x,int y)
 21 {
 22     return x>y? y:x;
 23 }
 24 bool fun(node &t)
 25 {
 26     int one,two;
 27     if(t.x>=1&&t.x<=n && t.y>=1&&t.y<=m && a[t.x][t.y]!='X')
 28     {
 29         one=abs(t.x-zx1)+abs(t.y-zy1);
 30         two=abs(t.x-zx2)+abs(t.y-zy2);
 31         one=(one+1)/2;
 32         two=(two+1)/2;
 33         one=Min(one,two);
 34         if(one>step) return false;
 35     }
 36     return true;
 37 }
 38 void bfs(int x)
 39 {
 40     int i,size1;
 41     node t,cur;
 42     size1=Q[x].size();
 43     while(size1--)
 44     {
 45         cur=Q[x].front();
 46         Q[x].pop();
 47         if(fun(cur))continue;
 48         for(i=0;i<4;i++)
 49         {
 50             t=cur;
 51             t.x=t.x+to[i][0];
 52             t.y=t.y+to[i][1];
 53             if(fun(t))continue;
 54             if(hash[x][t.x][t.y])continue;
 55             hash[x][t.x][t.y]=true;
 56             if(hash[x^1][t.x][t.y])
 57             {
 58                 glag=true;
 59                 return;
 60             }
 61             Q[x].push(t);
 62         }
 63     }
 64 }
 65 void dbfs()
 66 {
 67     node t;
 68     t.x=mx;
 69     t.y=my;
 70     hash[0][t.x][t.y]=true;
 71     Q[0].push(t);//boy 
 72     t.x=gx;
 73     t.y=gy;
 74     hash[1][t.x][t.y]=true;
 75     Q[1].push(t);//girl
 76     step=0;
 77     glag=false;
 78     while(true)
 79     {
 80         if(Q[0].size()==0 && Q[1].size()==0)break;
 81         step++;
 82         bfs(0);
 83         bfs(0);
 84         bfs(0);
 85         bfs(1);
 86         if(glag==true)break;
 87     }
 88     if(glag==true)printf("%d\n",step);
 89     else printf("-1\n");
 90 }
 91 void solve()
 92 {
 93     memset(hash,false,sizeof(hash));
 94     while(!Q[0].empty()){
 95         Q[0].pop();
 96     }
 97     while(!Q[1].empty()){
 98         Q[1].pop();
 99     }
100     dbfs();
101 }
102 int main()
103 {
104     int T;
105     bool flag;
106     int i,j;
107     scanf("%d",&T);
108     while(T--)
109     {
110         scanf("%d%d",&n,&m);
111         for(i=1;i<=n;i++)
112             scanf("%s",a[i]+1);
113         for(i=1,flag=false;i<=n;i++)
114             for(j=1;j<=m;j++)
115             {
116                 if(a[i][j]=='Z' && !flag){
117                     zx1=i;
118                     zy1=j;
119                     flag=true;
120                 }
121                 else if(a[i][j]=='Z' && flag){
122                     zx2=i;
123                     zy2=j;
124                 }
125                 if(a[i][j]=='M'){
126                     mx=i;
127                     my=j;
128                 }
129                 if(a[i][j]=='G'){
130                     gx=i;
131                     gy=j;
132                 }
133             }
134         solve();
135     }
136     return 0;
137 }

 

 

转载于:https://www.cnblogs.com/tom987690183/p/3663633.html