Find a way
Time Limit: 3000/1000 MS
(Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 3226 Accepted
Submission(s): 1045
Problem Description
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.
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.
Input
The 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
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
Output
For 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
Author
yifenfei
Source
Recommend
bfs简单广搜题。入门级难度，但是因为第一次做，还是耗费了不少时间。
题意是Y和M要去KFC一起吃饭，@为KFC的位置，可以有多个，求出他们到达KFC耗费的最短时间（1个格子耗费11分钟）。
用两次bfs()，分别求出Y和M到各KFC的最短路，相加之后，遍历出最短的总时间。
第一次练习广搜的题，代码写的有些费力：
1 #include <iostream>
2 #include <queue>
3 #include <string.h>
4 using namespace std;
5 char a[201][201];
6 int step[201][201];
7 int m1[201][201];
8 int m2[201][201];
9 bool isw[201][201];
10 int dx[4] = {0,1,0,-1};
11 int dy[4] = {1,0,-1,0};
12 int n,m;
13 int ycurx,ycury;
14 int mcurx,mcury;
15 struct NODE{
16 int x;
17 int y;
18 };
19 int Min(int a,int b)
20 {
21 return a<b?a:b;
22 }
23 bool judge(int x,int y)
24 {
25 if( x<1 || y<1 || x>n || y>m ) //越界
26 return 1;
27 if( isw[x][y] ) //走过了
28 return 1;
29 if( a[x][y]==‘#‘ ) //碰到墙
30 return 1;
31 return 0;
32 }
33 void bfs(int curx,int cury)
34 {
35 queue <NODE> q; //创建队列
36 NODE cur,next;
37 cur.x = curx;
38 cur.y = cury;
39 q.push(cur); //入队
40 memset(isw,0,sizeof(isw));
41 isw[curx][cury] = true;
42 step[curx][cury] = 0;
43 while(!q.empty()){
44 cur = q.front();
45 q.pop();
46 for(int i=0;i<4;i++){
47 int nx = cur.x + dx[i];
48 int ny = cur.y + dy[i];
49 if( judge(nx,ny) )
50 continue;
51 step[nx][ny] = step[cur.x][cur.y] + 1; //记录走到下一步的步数
52 isw[nx][ny] = true;
53 next.x = nx;
54 next.y = ny;
55 q.push(next);
56 }
57 }
58 }
59 int main()
60 {
61 while(cin>>n>>m){
62 for(int i=1;i<=n;i++)
63 for(int j=1;j<=m;j++){
64 cin>>a[i][j];
65 if(a[i][j]==‘Y‘){ //记录Y的位置
66 ycurx=i;
67 ycury=j;
68 }
69 else if(a[i][j]==‘M‘){ //记录M的位置
70 mcurx=i;
71 mcury=j;
72 }
73 }
74 int tmin = 999999999;
75 bfs(ycurx,ycury); //以（ycurx，ycury）为起点开始广度优先搜索
76 for(int i=1;i<=n;i++)
77 for(int j=1;j<=m;j++){
78 m1[i][j]=step[i][j];
79 }
80 bfs(mcurx,mcury); //以（mcurx，mcury）为起点开始广搜优先搜索
81 for(int i=1;i<=n;i++)
82 for(int j=1;j<=m;j++){
83 m2[i][j]=step[i][j];
84 }
85 for(int i=1;i<=n;i++) //遍历出最短的总步数
86 for(int j=1;j<=m;j++){
87 step[i][j]=m1[i][j] + m2[i][j];
88 if(a[i][j]==‘@‘){
89 tmin = Min(tmin,step[i][j]);
90 }
91 }
92 cout<<tmin*11<<endl; //输出最短总时间
93 }
94 return 0;
95 }
Freecode : www.cnblogs.com/yym2013