zoukankan      html  css  js  c++  java

  • HDU 1242 Rescue

    Rescue

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 32983    Accepted Submission(s): 11542


    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
     
     
    注意：可能有多个r
    AC代码(BFS+QUEUE)
    版本：a往多个r方向
     1 #include<iostream>
 2 #include<fstream>
 3 #include<queue>
 4 #include<string>
 5 #include<cstring>
 6 #include<vector>
 7 #include<limits>
 8 using namespace std;
 9 const int MAX=203;
10 struct pot
11 {
12     int x,y;
13 };
14 struct data
15 {
16     int time; char ch;
17 };
18 data arr[MAX][MAX];
19 int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
20 int n,m;
21 int anglex,angley;
22 int BFS()
23 {
24     int mintime=INT_MAX,flag=0;
25     pot beg={anglex,angley}; 
26     queue<pot> que;
27     que.push(beg);
28     while(!que.empty())
29     {
30         pot cur=que.front();
31         que.pop();
32         if(arr[cur.x][cur.y].ch=='r')
33         {    
34             if(arr[cur.x][cur.y].time<mintime)
35             {
36                 flag=1;
37                 mintime=arr[cur.x][cur.y].time;
38             }
39             //cout<<"sdsd:: "<<cur.x<<","<<cur.y<<"  :tie  "<<arr[cur.x][cur.y].time<<endl;
40             continue;
41         }
42         for(int i=0;i<4;i++)
43         {
44             pot next={cur.x+dir[i][0],cur.y+dir[i][1]};
45             if(next.x>=0&&next.x<n&&next.y>=0&&next.y<m&&arr[next.x][next.y].ch!='#')
46             {
47                 int zhi;
48                 if(arr[next.x][next.y].ch=='x') zhi=2;
49                 else zhi=1;
50                 if(flag==1)
51                 {
52                     if(arr[cur.x][cur.y].time+zhi<mintime&&(arr[next.x][next.y].time==0||arr[next.x][next.y].time>arr[cur.x][cur.y].time+zhi))
53                     {
54                         arr[next.x][next.y].time=arr[cur.x][cur.y].time+zhi;
55                         que.push(next);
56                     }
57                 }
58                 else
59                 {
60                     if(arr[next.x][next.y].time==0||arr[next.x][next.y].time>arr[cur.x][cur.y].time+zhi)
61                     {
62                         arr[next.x][next.y].time=arr[cur.x][cur.y].time+zhi;
63                         que.push(next);
64                     }
65                 }
66             }
67         }
68     }
69     return mintime;
70 }
71 int main()
72 {
73     //ifstream in("data.txt");
74     while(cin>>n>>m)
75     {
76         int i,j;
77         for(i=0;i<n;i++)
78             for(j=0;j<m;j++)
79             {
80                 cin>>arr[i][j].ch;
81                 arr[i][j].time=0;
82                 if(arr[i][j].ch=='a')
83                 { anglex=i; angley=j;}
84             }
85         int result=BFS();
86         if(result<INT_MAX)
87             cout<<result<<endl;
88         else
89             cout<<"Poor ANGEL has to stay in the prison all his life."<<endl;
90     }
91     return 0;
92 }

    AC代码(BFS+QUEUE)

    版本：多个r往a方向BFS

      1 #include<iostream>
  2 #include<fstream>
  3 #include<queue>
  4 #include<string>
  5 #include<cstring>
  6 #include<vector>
  7 #include<limits>
  8 using namespace std;
  9 const int MAX=203;
 10 struct pot
 11 {
 12     int x,y;
 13 };
 14 struct data
 15 {
 16     int time; char ch;
 17 };
 18 data arr[MAX][MAX];
 19 vector<pot> vec;
 20 int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
 21 int n,m;
 22 int anglex,angley;
 23 int BFS()
 24 {
 25     int j=-1;
 26     int mintime=INT_MAX,flag=0;
 27     while(++j<vec.size())
 28     {
 29         pot beg=vec[j]; 
 30         queue<pot> que;
 31         que.push(beg);
 32         while(!que.empty())
 33         {
 34             pot cur=que.front();
 35             que.pop();
 36             if(cur.x==anglex&&cur.y==angley)
 37             {    
 38                 if(arr[cur.x][cur.y].time<mintime)
 39                 {
 40                     flag=1;
 41                     mintime=arr[cur.x][cur.y].time;
 42                 }
 43                 continue;
 44             }
 45             for(int i=0;i<4;i++)
 46             {
 47                 pot next={cur.x+dir[i][0],cur.y+dir[i][1]};
 48                 if(next.x>=0&&next.x<n&&next.y>=0&&next.y<m&&arr[next.x][next.y].ch!='#'&&arr[next.x][next.y].ch!='r')
 49                 {
 50                     int zhi;
 51                     if(arr[next.x][next.y].ch=='.'||arr[next.x][next.y].ch=='a') zhi=0;
 52                     else zhi=1;
 53                     if(flag==1)
 54                     {
 55                         if(arr[cur.x][cur.y].time+zhi+1<mintime&&(arr[next.x][next.y].time==0||arr[next.x][next.y].time>arr[cur.x][cur.y].time+zhi+1))
 56                         {
 57                             arr[next.x][next.y].time=arr[cur.x][cur.y].time+zhi+1;
 58                             que.push(next);
 59                         }
 60                     }
 61                     else
 62                     {
 63                         if(arr[next.x][next.y].time==0||arr[next.x][next.y].time>arr[cur.x][cur.y].time+zhi+1)
 64                         {
 65                             arr[next.x][next.y].time=arr[cur.x][cur.y].time+zhi+1;
 66                             que.push(next);
 67                         }
 68                     }
 69                 }
 70             }
 71         }
 72     }
 73     return mintime;
 74 }
 75 int main()
 76 {
 77     //ifstream in("data.txt");
 78     while(cin>>n>>m)
 79     {
 80         int i,j;
 81         for(i=0;i<n;i++)
 82             for(j=0;j<m;j++)
 83             {
 84                 cin>>arr[i][j].ch;
 85                 arr[i][j].time=0;
 86                 if(arr[i][j].ch=='a')
 87                 { anglex=i; angley=j;continue;}
 88                 if(arr[i][j].ch=='r')
 89                 {
 90                     pot fri={i,j};
 91                     vec.push_back(fri);
 92                 }    
 93             }
 94         int result=BFS();
 95         if(result<INT_MAX)
 96             cout<<result<<endl;
 97         else
 98             cout<<"Poor ANGEL has to stay in the prison all his life."<<endl;
 99         vec.clear();
100     }
101     return 0;
102 }

    AC代码(BFS+优先队列)

    版本：从a开始BFS

     1 #include<iostream>
 2 #include<fstream>
 3 #include<queue>
 4 #include<string>
 5 #include<cstring>
 6 using namespace std;
 7 const int MAX=203;
 8 struct pot
 9 {
10     int x,y;
11 };
12 struct data
13 {
14     int x,y,time;
15     friend bool operator<(const data &a,const data &b)
16     {  
17         return a.time>b.time;  
18     }  
19 };
20 char arr[MAX][MAX];
21 int vis[MAX][MAX];
22 int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
23 int n,m;
24 int anglex,angley;
25 int BFS()
26 {
27     data beg={anglex,angley,0}; 
28     priority_queue<data> que;
29     que.push(beg);
30     vis[anglex][angley]=1;
31     while(!que.empty())
32     {
33         data cur=que.top();
34         que.pop();
35         if(arr[cur.x][cur.y]=='r')
36             return cur.time;
37         for(int i=0;i<4;i++)
38         {
39             data next={cur.x+dir[i][0],cur.y+dir[i][1],0};
40             if(next.x>=0&&next.x<n&&next.y>=0&&next.y<m&&arr[next.x][next.y]!='#'&&vis[next.x][next.y]==0)
41             {
42                 if(arr[next.x][next.y]=='x')
43                     next.time=cur.time+2;
44                 else
45                     next.time=cur.time+1;
46                 vis[next.x][next.y]=1;
47                 que.push(next);
48             }
49         }
50     }
51     return 0;
52 }
53 int main()
54 {
55     //ifstream in("data.txt");
56     while(cin>>n>>m)
57     {
58         int i,j;
59         for(i=0;i<n;i++)
60             for(j=0;j<m;j++)
61             {
62                 cin>>arr[i][j];
63                 if(arr[i][j]=='a')
64                 { anglex=i; angley=j;}
65             }
66         memset(vis,0,sizeof(vis));
67         int result=BFS();
68         if(result>0)
69             cout<<result<<endl;
70         else
71             cout<<"Poor ANGEL has to stay in the prison all his life."<<endl;
72     }
73     return 0;
74 }
    //a往r方向BFS
  • 相关阅读:
    【心得】软件团队Git工作流及Jira的基本知识和常见问题解释
    项目系统Netty的Channel和用户之间的关系绑定正确做法,以及Channel通道的安全性方案
    Redis中的事务(多命令)操作
    Redis中的订阅模式
    Redis中有序列表(ZSet)相关命令
    Redis散列(Hash)的相关命令
    输入输出流String间的转换
    linux 常用命令
    Gradle 使用
    c 学习笔记 0
  • 原文地址：https://www.cnblogs.com/zhaopeng938/p/7921397.html
Copyright © 2011-2022 走看看