HDU 2579.Dating with girls(2)-白红宇

HDU 2579.Dating with girls(2)

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发布时间：2019-06-06

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Dating with girls(2)

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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Description

If you have solved the problem Dating with girls(1).I think you can solve this problem too.This problem is also about dating with girls. Now you are in a maze and the girl you want to date with is also in the maze.If you can find the girl, then you can date with the girl.Else the girl will date with other boys. What a pity!

The Maze is very strange. There are many stones in the maze. The stone will disappear at time t if t is a multiple of k(2<= k <= 10), on the other time , stones will be still there.

There are only ‘.’ or ‘#’, ’Y’, ’G’ on the map of the maze. ’.’ indicates the blank which you can move on, ‘#’ indicates stones. ’Y’ indicates the your location. ‘G’ indicates the girl's location . There is only one ‘Y’ and one ‘G’. Every seconds you can move left, right, up or down.

Input

The first line contain an integer T. Then T cases followed. Each case begins with three integers r and c (1 <= r , c <= 100), and k(2 <=k <= 10).

The next r line is the map’s description.

Output

For each cases, if you can find the girl, output the least time in seconds, else output "Please give me another chance!".

Sample Input

1 6 6 2 ...Y.. ...#.. .#.... ...#.. ...#.. ..#G#.

Sample Output

7

由于在不同时间，墙有存在和不存在两种情况。

因此，在判重时，也要考虑到时间。只需考虑time%k即可

其他部分就是正常的BFS问题

1 /*  2 By:OhYee  3 Github:OhYee  4 Email:oyohyee@oyohyee.com  5 Blog:http://www.cnblogs.com/ohyee/  6   7 かしこいかわいい？  8 エリーチカ！  9 要写出来Хорошо的代码哦~ 10 */ 11  12 #include 
          
            13 #include 
           
             14 #include 
            
              15 #include 
             
               16 #include 
              
                17 #include 
               
                 18 #include 
                
                  19 #include 
                 
                   20 #include 
                  
                    21 #include 
                   
                     22 #include 
                     23 using namespace std; 24 25 //DEBUG MODE 26 #define debug 0 27 28 //循环 29 #define REP(n) for(int o=0;o
                     
                       Q; 52 53 Q.push(point(s1,s2,0)); 54 dis[s1][s2][0] = 0; 55 while(!Q.empty()) { 56 int x = Q.front().x; 57 int y = Q.front().y; 58 int dist = Q.front().dis; 59 Q.pop(); 60 61 REP(4) { 62 int xx = x + delta[o]; 63 int yy = y + delta[3 - o]; 64 if(xx < 0 || xx >= n || yy < 0 || yy >= m) 65 continue; 66 if(Map[xx][yy] != '#' || (Map[xx][yy] == '#' && (dist+1) % k == 0)) { 67 int dd = dist + 1; 68 if(dis[xx][yy][dd%k] == -1) { 69 dis[xx][yy][dd%k] = dd; 70 if(xx == v1 && yy == v2) 71 return dd; 72 Q.push(point(xx,yy,dd)); 73 74 } 75 } 76 } 77 } 78 return -1; 79 } 80 81 void Do() { 82 scanf("%d%d%d",&n,&m,&k); 83 int s1,s2,v1,v2; 84 for(int i = 0;i < n;i++) 85 for(int j = 0;j < m;j++) { 86 scanf("\n%c\n",&Map[i][j]); 87 if(Map[i][j] == 'Y') { 88 s1 = i; 89 s2 = j; 90 } 91 if(Map[i][j] == 'G') { 92 v1 = i; 93 v2 = j; 94 } 95 } 96 int ans = BFS(s1,s2,v1,v2); 97 if(ans == -1) 98 printf("Please give me another chance!\n"); 99 else100 printf("%d\n",ans);101 }102 103 int main() {104 int T;105 scanf("%d",&T);106 while(T--)107 Do();108 return 0;109 }