hdu1242 Rescue(BFS)

hdu1242 Rescue(BFS)

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 21701    Accepted Submission(s): 7745

Problem Description

Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison. Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards. You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)

Input

First line contains two integers stand for N and M. Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.  Process to the end of the file.

Output

For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."

Sample Input

7 8 #.#####. #.a#..r. #..#x... ..#..#.# #...##.. .#...... ........

Sample Output

13

Author

CHEN, Xue

Source

​​ZOJ Monthly, October 2003​​

Recommend

Eddy   |   We have carefully selected several similar problems for you:   ​​1240​​​  ​​​1010​​​  ​​​1072​​​  ​​​1253​​​  ​​​1372​​

​​Statistic​​ |

​​Submit​​ |

​​Discuss​​ |

​​Note​​

也可以用DFS做 在这里用的BFS

#include #include #include using namespace std;struct node{ int cost,x,y; friend bool operator<(node a,node b) { return a.cost>b.cost; }};int dir[4][2]={0,1,0,-1,1,0,-1,0};char map[205][205];int vis[205][205],n,m;bool limit(int x,int y){ if(map[x][y]!='#'&&x>=0&&x=0&&ys; node temp,temp1; int flag=0; temp.x=x,temp.y=y,temp.cost=0; s.push(temp); while(!s.empty()) { temp1=temp=s-(); s.pop(); if(map[temp.x][temp.y]=='a') { flag=1; break; } for(int i=0;i<4;i++) { temp.x=temp.x+dir[i][0]; temp.y=temp.y+dir[i][1]; if(!vis[temp.x][temp.y]&&limit(temp.x,temp.y)) { vis[temp.x][temp.y]=1; temp.cost+=tonum(temp.x,temp.y); s.push(temp); } temp=temp1; } } if(flag) printf("%d\n",temp.cost); else printf("Poor ANGEL has to stay in the prison all his life.\n"); while(!s.empty()) s.pop();}int main(){ int st_x,st_y; while(scanf("%d %d",&n,&m)!=EOF) { for(int i=0;i

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。