java写bfs

本文介绍了一个算法问题,旨在找到两个角色从各自位置出发,到达宁波城市中多个肯德基门店之一的最短总时间。通过使用广度优先搜索算法,文章详细解释了如何在给定地图上找到最佳会面点,同时考虑到移动限制和目标位置。

Find a way

 HDU - 2612

Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki. 
Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest. 
Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes. 

InputThe input contains multiple test cases. 
Each test case include, first two integers n, m. (2<=n,m<=200). 
Next n lines, each line included m character. 
‘Y’ express yifenfei initial position. 
‘M’    express Merceki initial position. 
‘#’ forbid road; 
‘.’ Road. 
‘@’ KCF 
OutputFor each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.Sample Input

4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#

Sample Output

66
88
66
  1 import java.util.ArrayDeque;
  2 import java.util.PriorityQueue;
  3 import java.util.Scanner;
  4 class node{
  5     int x;
  6     int y;
  7     int step;
  8     int flag;
  9 }
 10 public class Main {
 11     static int n,m;
 12     static int [][][] vis = new int[220][220][2];
 13     static char[][] c = new char[220][220];
 14     static int [][][] dis = new int[220][220][2];
 15     static String[] s = new String[220];
 16     static int [][] dir = {{1,0},{0,1},{-1,0},{0,-1}};
 17     static node head,tail;
 18     static ArrayDeque<node> q = new ArrayDeque<node>();
 19     static void bfs() {
 20         while(!q.isEmpty()) {
 21             head = q.poll();
 22             //System.out.println(head.x+" "+head.y+" "+head.flag);
 23             //System.out.println("lallaa"+vis[head.x][head.y][head.flag]);
 24             for(int i=0;i<4;i++) {
 25                 int tx = head.x+dir[i][0];
 26                 int ty = head.y+dir[i][1];
 27                 int tstep = head.step + 1;
 28                 if(tx<0||tx>=n||ty<0||ty>=m||c[tx][ty]=='#'||vis[tx][ty][head.flag]==1)
 29                     continue;
 30                 if(c[tx][ty]=='@') {
 31                     if(head.flag==0) {
 32                     //    System.out.println("haah");
 33                         dis[tx][ty][0] = Math.min(dis[tx][ty][0], tstep);
 34                         vis[tx][ty][0] = 1;
 35                     }
 36                     else if(head.flag==1) {
 37                         dis[tx][ty][1] = Math.min(dis[tx][ty][1], tstep);
 38                         vis[tx][ty][1] = 1;
 39                     }
 40                 }
 41                 vis[tx][ty][head.flag] = 1;
 42                 tail = new node();
 43                 tail.flag = head.flag;
 44                 tail.step = head.step + 1;
 45                 tail.x = tx;
 46                 tail.y = ty;
 47                 q.offer(tail);
 48             }
 49         }
 50     }
 51     public static void main(String[] args) {
 52         Scanner cin = new Scanner(System.in);
 53         while(cin.hasNext()) {
 54             n = cin.nextInt();
 55             m = cin.nextInt();
 56             for(int i=0;i<n;i++) {
 57                 for(int j=0;j<m;j++) {
 58                     for(int k=0;k<2;k++) {
 59                         dis[i][j][k] = 1000000;
 60                     }
 61                 }
 62             }
 63             for(int i=0;i<n;i++) {
 64                 for(int j=0;j<m;j++) {
 65                     for(int k=0;k<2;k++) {
 66                         vis[i][j][k] = 0;
 67                     }
 68                 }
 69             }
 70             for(int i=0;i<n;i++) {
 71                 s[i] = cin.next();
 72             }
 73             for(int i=0;i<n;i++) {
 74                 for(int j=0;j<s[i].length();j++) {
 75                     c[i][j] = s[i].charAt(j);
 76                     if(c[i][j] == 'Y') {
 77                         head = new node();
 78                         head.step = 0;
 79                         head.flag = 0;
 80                         head.x = i;
 81                         head.y = j;
 82                         q.offer(head);
 83                         vis[i][j][0] = 1;
 84                         //bfs();
 85                     }
 86                     if(c[i][j] == 'M') {
 87                         head = new node();
 88                         head.step = 0;
 89                         head.flag = 1;
 90                         head.x = i;
 91                         head.y = j;
 92                         q.offer(head);
 93                         vis[i][j][1] = 1;
 94                         
 95                     }
 96                 }
 97             }
 98             bfs();
 99             int ans = 1000000;
100             for(int i=0;i<n;i++) {
101                 for(int j=0;j<m;j++) {
102                     if(c[i][j] == '@') {
103                         ans = Math.min(ans,dis[i][j][0]+dis[i][j][1]);
104                     }
105                 }
106             }
107             System.out.println(ans*11);
108         }
109     }
110 }

 

转载于:https://www.cnblogs.com/1013star/p/10353589.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值