Problem Description The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze was changed and the way he came in was lost.He realized that the bone was a trap, and he tried desperately to get out of this maze.
The maze was a rectangle with the sizes of N by M. The maze is made up of a door,many walls and many explosives. Doggie need to reach the door to escape from the tempter. In every second, he could move one block to one of the upper, lower, left or right neighboring blocks. And if the destination is a wall, Doggie need another second explode and a explosive to explode it if he had some explosives. Once he entered a block with explosives,he can take away all of the explosives. Can the poor doggie survive? Please help him.
Input The input consists of multiple test cases. The first line of each test case contains two integers N, M,(2 <= N, M <= 8). which denote the sizes of the maze.The next N lines give the maze layout, with each line containing M characters. A character is one of the following:
'X': a block of wall;
'S': the start point of the doggie;
'D': the Door;
'.': an empty block;
'1'--'9':explosives in that block.
Note,initially he had no explosives.
The input is terminated with two 0's. This test case is not to be processed.
Output For each test case, print the minimum time the doggie need to escape if the doggie can survive, or -1 otherwise.
Sample Input
4 4
SX..
XX..
....
1..D
4 4
S.X1
....
..XX
..XD
0 0
Sample Output
-1
9
Author XHD
Source HDU 2007-10 Programming Contest
思路：因为地图会变化，结构体里面要加一个数组保存地图，标记的话就标记根据地图的坐标和当前手中的炸药数量来标记，但是同样状态的标记可能会有不同的走法，所以用整形变量来存，并且设定一个状态可以访问的阀值，这里设为20。
#include
#include
#include
using namespace std; struct S{ int x,y,num,step; char mp[8][8]; bool operator<(const S & p) const { return step>p.step; } }t; int nxt[4][2]={{1,0},{0,1},{-1,0},{0,-1}}; char s[8][9]; int vis[8][8][577]; int main() { int n,m,i,j,sx,sy,ex,ey; while(~scanf("%d%d",&n,&m) && n) { priority_queue
que; for(i=0;i
=0 && t.x
=0 && t.y
0) { vis[t.x][t.y][t.num]++; t.step++; t.num--; t.mp[t.x][t.y]='.'; vis[t.x][t.y][t.num]++; que.push(t); } else if(t.mp[t.x][t.y]>='1' && t.mp[t.x][t.y]<='9') { vis[t.x][t.y][t.num]++; t.num+=t.mp[t.x][t.y]-'0'; t.mp[t.x][t.y]='.'; vis[t.x][t.y][t.num]++; que.push(t); } } t=que.top(); } que.pop(); } if(que.empty()) printf("-1\n"); } }