HDU-2612-Find a way

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2612

Find a way

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.

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

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 4Y.#@.....#..@..M4 4Y.#@.....#..@#.M5 5Y..@..#....#...@..M.#...#

Sample Output

668866

 题目分析：Y和M两个人从各自的起点出发，约定在kfc见面，求二人到kfc的步数之和最小。经典bfs。

#include<iostream>#include<queue>#include<cstring>using namespace std;int n,m;int ans1[201][201],ans2[201][201];int vis1[201][201],vis2[201][201];int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};char mp[201][201];struct node{    int x,y;};bool check(node t){    if(t.x>=0&&t.x<n&&t.y>=0&&t.y<m&&mp[t.x][t.y]!='#')        return 1;    else        return 0;}void bfs(int sx,int sy,int vis[201][201],int ans[201][201]){    queue<node>q;    node now,next;    now.x=sx;now.y=sy;    vis[now.x][now.y]=1;    ans[now.x][now.y]=0;    q.push(now);while(!q.empty())    {        now=q.front();        q.pop();        for(int i=0;i<4;i++)        {            next.x=now.x+dir[i][0];            next.y=now.y+dir[i][1];            if(check(next)&&!vis[next.x][next.y])            {                vis[next.x][next.y]=1;                ans[next.x][next.y]=ans[now.x][now.y]+1;                q.push(next);            }        }    } }int main(){    int x1,y1,x2,y2;    while(cin>>n>>m)    {    if(n==0&&m==0)    break;    for(int i=0;i<n;i++)    {    for(int j=0;j<m;j++)    {    cin>>mp[i][j];    if(mp[i][j]=='Y')                {                  x1=i;                  y1=j;                }                else if(mp[i][j]=='M')                {                  x2=i;                  y2=j;                }}}int min=10000;memset(vis1,0,sizeof(vis1));memset(ans1,0,sizeof(ans1));bfs(x1,y1,vis1,ans1);memset(vis2,0,sizeof(vis2));memset(ans2,0,sizeof(ans2));bfs(x2,y2,vis2,ans2);for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(mp[i][j]=='@'&&vis1[i][j]&&vis2[i][j])//这个点必须被同时访问到。 {if(ans1[i][j]+ans2[i][j]<min){min=ans1[i][j]+ans2[i][j];}}}}cout<<min*11<<endl;}    return 0;}