HDU 1242 Rescue

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38663    Accepted Submission(s): 13362

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

 

 

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采用bfs搜索算法

但是采用bfs的时候要注意遇到守卫的情况，应该延后考虑

如果守卫先进入队列

那么周围的会被标记为已读

影响周围更优解法的生成

所以要延后遇到守卫时的优先级顺序

下面上我有注释

而且谁都能看懂的代码

 

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  1 //Author:LanceYu
  2 #include<iostream>
  3 #include<string>
  4 #include<cstring>
  5 #include<cstdio>
  6 #include<fstream>
  7 #include<iosfwd>
  8 #include<sstream>
  9 #include<fstream>
 10 #include<cwchar>
 11 #include<iomanip>
 12 #include<ostream>
 13 #include<vector>
 14 #include<cstdlib>
 15 #include<queue>
 16 #include<set>
 17 #include<ctime>
 18 #include<algorithm>
 19 #include<complex>
 20 #include<cmath>
 21 #include<valarray>
 22 #include<bitset>
 23 #include<iterator>
 24 #define ll long long
 25 using namespace std;
 26 const double clf=1e-8;
 27 //const double e=2.718281828;
 28 const double PI=3.141592653589793;
 29 const int MMAX=2147483647;
 30 //priority_queue<int>p;
 31 //priority_queue<int,vector<int>,greater<int> >pq;
 32 int Min,n,m,dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//四个方向 
 33 char map[201][201];
 34 struct node
 35 {
 36     int x,y,step;
 37 };
 38 int bfs(int x,int y,int x1,int y1)
 39 {
 40     Min=MMAX;
 41     int i;
 42     queue<node> q;
 43     q.push(node{x,y,0});
 44     map[x][y]='#';
 45     while(!q.empty())
 46     {
 47         node t=q.front();
 48         q.pop();
 49         if(t.x==x1&&t.y==y1)
 50             return t.step;
 51         if(map[t.x][t.y]=='X')//当发现之前是守卫的点，标记为不可走，步数+1，直接走到下一步 
 52         {
 53             map[t.x][t.y]='#';
 54             t.step++;
 55             q.push(node{t.x,t.y,t.step});
 56             continue;
 57         }
 58         for(i=0;i<4;i++)
 59         {
 60             int dx=t.x+dir[i][0];
 61             int dy=t.y+dir[i][1];
 62             if(dx>=0&&dy>=0&&dx<m&&dy<n&&map[dx][dy]=='a')//如果是终点或者正常路径，则标记为不可读，入列 
 63             {
 64                 map[dx][dy]='#';
 65                 q.push(node{dx,dy,t.step+1});
 66             }
 67             if(dx>=0&&dy>=0&&dx<m&&dy<n&&map[dx][dy]=='.')
 68             {
 69                 map[dx][dy]='#';
 70                 q.push(node{dx,dy,t.step+1});
 71             if(dx>=0&&dy>=0&&dx<m&&dy<n&&map[dx][dy]=='x')//用X作为标记，使其优先级移到队列最后 
 72             {
 73                 map[dx][dy]='X';
 74                 q.push(node{dx,dy,t.step+1});
 75                 
 76             }
 77             
 78             }
 79         }
 80     }
 81     return 0;//找不到返回0 
 82 }
 83 int main()
 84 {
 85     int a,b,a1,b1;
 86     while(scanf("%d%d",&m,&n)!=EOF)
 87     {
 88         for(int i=0;i<m;i++)
 89         {
 90             scanf("%s",map[i]);
 91             for(int j=0;j<n;j++)
 92             {
 93                 if(map[i][j]=='r')//找起点 
 94                 {
 95                     a=i;
 96                     b=j;
 97                 }
 98                 if(map[i][j]=='a')//找终点 
 99                 {
100                     a1=i;
101                     b1=j;
102                 }
103             }
104         }
105         int ans=bfs(a,b,a1,b1);
106         if(!ans)//如果ans不等于0就输出ans 
107             printf("Poor ANGEL has to stay in the prison all his life.\n");
108         else
109             printf("%d\n",ans);
110     }
111     return 0;
112 }

2018-11-15　　23:49:18　　Author:LanceYu