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
营救天使行动。
不同于一般的广搜记录步数，这里是记录的最短时间。
同时需要进行剪枝，当到达一个点时，如果该点原先的时间要比更新后的时间长，则入队，否则不入队。
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string.h> 4 #include<math.h> 5 #include<algorithm> 6 #include<queue> 7 #include<stack> 8 #include<deque> 9 #include<iostream> 10 using namespace std; 11 typedef long long LL; 12 const double pi=acos(-1.0); 13 const double e=exp(1); 14 15 struct con{ 16 int x; 17 int y; 18 }; 19 int Map[210][210]; 20 int n,m,t; 21 int check[50000]; 22 int times[210][210]; 23 int way[4][2]={0,1,0,-1,1,0,-1,0}; 24 25 void BFS(int a,int b) 26 { 27 int i,p,j,min1=99999999; 28 queue<struct con> qq; 29 struct con mid,head; 30 head.x=a; 31 head.y=b; 32 // check[Map[head.x][head.y]]=1; 33 qq.push(head); 34 while(!qq.empty()) 35 { 36 head=qq.front(); 37 qq.pop(); 38 if(Map[head.x][head.y]==-1) 39 { 40 if(times[head.x][head.y]<min1) 41 min1=times[head.x][head.y]; 42 } 43 for(i=0;i<=3;i++) 44 { 45 mid.x=head.x+way[i][0]; 46 mid.y=head.y+way[i][1]; 47 48 if(mid.x>=0&&mid.y>=0&&mid.x<n&&mid.y<m) 49 { 50 if(Map[mid.x][mid.y]==0&×[head.x][head.y]+2<times[mid.x][mid.y]) 51 { 52 times[mid.x][mid.y]=times[head.x][head.y]+2; 53 qq.push(mid); 54 } 55 else if(Map[mid.x][mid.y]==-2) 56 ; 57 else if(Map[mid.x][mid.y]!=0&×[head.x][head.y]+1<times[mid.x][mid.y]) 58 { 59 times[mid.x][mid.y]=times[head.x][head.y]+1; 60 qq.push(mid); 61 } 62 } 63 } 64 } 65 if(min1==99999999) 66 printf("Poor ANGEL has to stay in the prison all his life. "); 67 else 68 printf("%d ",min1); 69 return ; 70 } 71 int main() 72 { 73 int i,p,j; 74 char c; 75 int a,b; 76 while(scanf("%d%d",&n,&m)!=EOF) 77 { 78 p=1; 79 memset(check,0,sizeof(check)); 80 memset(times,0x3f3f3f3f,sizeof(times)); 81 for(i=0;i<n;i++) 82 for(j=0;j<m;j++) 83 { 84 scanf(" %c",&c); 85 if(c=='r') 86 { 87 a=i; 88 b=j; 89 Map[i][j]=p++; 90 times[i][j]=0; 91 } 92 else if(c=='a') 93 Map[i][j]=-1; 94 else if(c=='#') 95 Map[i][j]=-2; 96 else if(c=='x') 97 Map[i][j]=0; 98 else if(c=='.') 99 Map[i][j]=p++; 100 } 101 102 BFS(a,b); 103 } 104 return 0; 105 }