题目描述：
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.
输入：
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
输出：
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.
样例输入：
4 4 Y.#@ .... .#.. @..M 4 4 Y.#@ .... .#.. @#.M 5 5 Y..@. .#... .#... @..M. #...#
样例输出：
66 88 66
这题的意思就是两个人从两个不同的地点去一个肯德基见面，但是肯德基有很多个，要找出两个人花费时间最少的那一个，并输出这个时间，走一步用11min。
建立一个数组记录到各个点的时间是多少，最后遍历一遍，找出这个地方是肯德基并且花费时间最少的。
C++代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=205;
const int INF=0x7fffffff;
char ch[maxn][maxn];
int vis[maxn][maxn];
int step[maxn][maxn];
int n,m;
int dis[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
struct node
{
int x,y,tim;
};
void bfs(int a,int b)
{
queue<node>q;
node s,e;
s.x=a;
s.y=b;
s.tim=0;
vis[s.x][s.y]=1;
q.push(s);
while(!q.empty())
{
s=q.front();
q.pop();
for(int i=0;i<4;i++)
{
e.x=s.x+dis[i][0];
e.y=s.y+dis[i][1];
if(ch[e.x][e.y]=='#'||e.x<0||e.y<0||e.x>=n||e.y>=m||vis[e.x][e.y])
continue;
e.tim=s.tim+1;
step[e.x][e.y]+=e.tim;
vis[e.x][e.y]=1;
q.push(e);
}
}
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
int minn=INF;
memset(ch,0,sizeof(ch));
memset(step,0,sizeof(step));
for(int i=0;i<n;i++)
scanf("%s",ch[i]);
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(ch[i][j]=='Y'||ch[i][j]=='M')
{
memset(vis,0,sizeof(vis));
bfs(i,j);
}
}
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(ch[i][j]=='@'&&step[i][j]!=0)
minn=min(minn,step[i][j]);
}
printf("%d\n",minn*11);
}
return 0;
}
Java代码：
版本1 分别用step数组表示Y和M到@的距离，最后相加再比较，step为三维数组。
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.math.*;
public class Main {
final static int maxn=205;
final static int INF=0x7fffffff;
public static int n,m;
public static int [][]dis={{0,1},{0,-1},{1,0},{-1,0}};
public static boolean[][] vis=new boolean [maxn][maxn];
public static int [][][]step=new int [2][maxn][maxn];
public static char [][]ch=new char [maxn][maxn];
public static class node
{
int x,y;
node()
{
}
node(int x,int y)
{
this.x=x;
this.y=y;
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner cin =new Scanner(System.in);
while(cin.hasNext())
{
for(int i = 0; i < 2; i++)
for(int j = 0; j < maxn; j++)
for(int k = 0; k < maxn; k++)
step[i][j][k] = 0;
int minn=INF;
n=cin.nextInt();
m=cin.nextInt();
for(int i=0;i<n;i++)
{
ch[i]=cin.next().toCharArray();
}
Queue<node>q =new LinkedList<>();
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(ch[i][j]=='Y'||ch[i][j]=='M')
{
for(int k=0;k<n;k++)
for(int d=0;d<m;d++)
vis[k][d]=false;
bfs(i,j, ch[i][j] == 'Y' ? 0 : 1);
}
}
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(ch[i][j]=='@'&&step[0][i][j]!=0&&step[1][i][j] != 0)
minn=Math.min(step[0][i][j] + step[1][i][j], minn);
}
System.out.println(minn);
}
}
private static void bfs(int x, int y, int cnt) {
// TODO Auto-generated method stub
Queue<node>q =new LinkedList<>();
vis[x][y]=true;
// step[cnt][x][y] = 0;
q.add(new node(x, y));
while(!q.isEmpty())
{
node s=q.peek();
q.poll();
//System.out.println(s.x + " " + s.y);
for(int i=0;i<4;i++)
{
node e = new node();
e.x=s.x+dis[i][0];
e.y=s.y+dis[i][1];
if(e.x<0||e.y<0||e.x>=n||e.y>=m || ch[e.x][e.y] == '#' || vis[e.x][e.y]) continue;
step[cnt][e.x][e.y] = step[cnt][s.x][s.y] + 11;
vis[e.x][e.y]=true;
q.add(e);
}
}
}
}
版本2 step数组直接表示Y和M到达@的总时间，step为二维数组。
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.math.*;
public class Main {
final static int maxn=205;
final static int INF=0x7fffffff;
public static int n,m;
public static int [][]dis={{0,1},{0,-1},{1,0},{-1,0}};
public static boolean[][] vis=new boolean [maxn][maxn];
public static int [][]step=new int [maxn][maxn];
public static char [][]ch=new char [maxn][maxn];
public static class node
{
int x,y,tim;
node()
{
}
node(int x,int y,int tim)
{
this.x=x;
this.y=y;
this.tim=tim;
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner cin =new Scanner(System.in);
while(cin.hasNext())
{
for(int j = 0; j < maxn; j++)
for(int k = 0; k < maxn; k++)
step[j][k] = 0;
int minn=INF;
n=cin.nextInt();
m=cin.nextInt();
for(int i=0;i<n;i++)
{
ch[i]=cin.next().toCharArray();
}
Queue<node>q =new LinkedList<>();
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(ch[i][j]=='Y'||ch[i][j]=='M')
{
for(int k=0;k<n;k++)
for(int d=0;d<m;d++)
vis[k][d]=false;
bfs(i,j);
}
}
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(ch[i][j]=='@'&&step[i][j]!=0)
minn=Math.min(step[i][j], minn);
}
System.out.println(minn*11);
}
}
private static void bfs(int x, int y) {
// TODO Auto-generated method stub
Queue<node>q =new LinkedList<>();
vis[x][y]=true;
q.add(new node(x, y,0));
while(!q.isEmpty())
{
node s=q.peek();
q.poll();
//System.out.println(s.x + " " + s.y);
for(int i=0;i<4;i++)
{
node e = new node();
e.x=s.x+dis[i][0];
e.y=s.y+dis[i][1];
if(e.x<0||e.y<0||e.x>=n||e.y>=m || ch[e.x][e.y] == '#' || vis[e.x][e.y]) continue;
e.tim=s.tim+1;
step[e.x][e.y] += e.tim;
vis[e.x][e.y]=true;
q.add(e);
}
}
}
}
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