HDU1242 BFS+优先队列

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27406    Accepted Submission(s): 9711

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

题意：

~

代码：

 //一道很简单的题卡了很长时间。从重点出发找起点，优先队列存
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<queue>
 #include<vector>
 #include<functional>
 using namespace std;
 int n,m;
 struct Lu
 {
     int x,y,cnt;
     friend bool operator<(Lu a,Lu b)
     {
         return a.cnt>b.cnt;
     }
 };
 bool vis[][];
 int dir[][]={,,-,,,,,-};
 char ch[][];
 int bfs(int sx,int sy)
 {
     priority_queue<Lu>q;
     Lu L1,L2;
     L1.x=sx;L1.y=sy;L1.cnt=;
     q.push(L1);
     vis[sx][sy]=;
     while(!q.empty())
     {
         L2=q.top();
         q.pop();
         if(ch[L2.x][L2.y]=='r')
         return L2.cnt;
         for(int i=;i<;i++)
         {
             L1.x=L2.x+dir[i][];
             L1.y=L2.y+dir[i][];
             if(L1.x<||L1.x>=n||L1.y<||L1.y>=m) continue;
             if(vis[L1.x][L1.y]) continue;
             if(ch[L1.x][L1.y]=='#') continue;
             if(ch[L1.x][L1.y]=='x')
             L1.cnt=L2.cnt+;
             else L1.cnt=L2.cnt+;
             vis[L1.x][L1.y]=;
             q.push(L1);
         }
     }
     return -;
 }
 int main()
 {
     int sx,sy;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         for(int i=;i<n;i++)
         {
             scanf("%s",ch[i]);
             for(int j=;j<m;j++)
             {
                 if(ch[i][j]=='a')
                 {
                     sx=i;sy=j;
                 }
             }
         }
         memset(vis,,sizeof(vis));
         int ans=bfs(sx,sy);
         if(ans==-) printf("Poor ANGEL has to stay in the prison all his life.\n");
         else printf("%d\n",ans);
     }
     return ;
 }