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 }
原文：https://www.cnblogs.com/1013star/p/10353589.html