HDU

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

Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16192    Accepted Submission(s): 5198

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

 

Author

yifenfei

 

Source

奋斗的年代

 

Recommend

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Y和M去一个肯德基会面 一定有一个肯德基能到达 

已Y和M为起点 对图广搜 book标记步数

将所有的kfc点存在一个容器中 然后依次判断最小值（注意这里可能存在有的kfc不能到达 这个地方栽跟头了 记得多加这个判断就行）

#include <iostream>#include <queue>#include <stdio.h>#include <stdlib.h>#include <stack>#include <limits.h>#include <string>#include <string.h>#include <vector>#include <set>#include <map>#include <algorithm>#include <math.h>using namespace std;char mmap[222][222];int bookm[222][222];int booky[222][222];struct xjy{    int x;    int y;};xjy bem,bey;int n,m,k;queue <xjy > q,qq;int dir[4][2]={1,0,-1,0,0,1,0,-1};void bfsm(){    while(!q.empty())        q.pop();    for(int i=0;i<=201;i++)        for(int j=0;j<=201;j++)    {        bookm[i][j]=INT_MAX;    }    bookm[bem.x][bem.y]=0;    xjy mid;    q.push(bem);    while(!q.empty())    {        mid=q.front();        q.pop();        xjy midmid;        for(int i=0;i<4;i++)        {            midmid.x=mid.x+dir[i][0];            midmid.y=mid.y+dir[i][1];            if(mmap[midmid.x][midmid.y]!='#'&&mmap[midmid.x][midmid.y]!='M'&&bookm[midmid.x][midmid.y]>(bookm[mid.x][mid.y]+1))            {                bookm[midmid.x][midmid.y]=bookm[mid.x][mid.y]+1;                q.push(midmid);            }        }    }}void bfsy(){    while(!q.empty())        q.pop();    for(int i=0;i<=201;i++)        for(int j=0;j<=201;j++)    {        booky[i][j]=INT_MAX;    }    booky[bey.x][bey.y]=0;    xjy mid;    q.push(bey);    while(!q.empty())    {        mid=q.front();        q.pop();        xjy midmid;        for(int i=0;i<4;i++)        {            midmid.x=mid.x+dir[i][0];            midmid.y=mid.y+dir[i][1];            if(mmap[midmid.x][midmid.y]!='#'&&mmap[midmid.x][midmid.y]!='Y'&&booky[midmid.x][midmid.y]>(booky[mid.x][mid.y]+1))            {                booky[midmid.x][midmid.y]=booky[mid.x][mid.y]+1;                q.push(midmid);            }        }    }}int main(){    while(~scanf("%d%d",&n,&m))    {        while(!qq.empty())            qq.pop();        memset(mmap,'#',sizeof(mmap));        for(int i=1;i<=n;i++)            for(int j=1;j<=m;j++)                {                    scanf(" %c",&mmap[i][j]);                    if(mmap[i][j]=='@')                    {                        xjy mid;                        mid.x=i;mid.y=j;                        qq.push(mid);                    }                    else if(mmap[i][j]=='Y')                    {                        bey.x=i;bey.y=j;                    }                    else if(mmap[i][j]=='M')                    {                        bem.x=i;bem.y=j;                    }                }    bfsm();    bfsy();    int ans=INT_MAX;    while(!qq.empty())    {        xjy mid;        mid=qq.front();        qq.pop();        int aans=INT_MAX;        if(booky[mid.x][mid.y]!=INT_MAX&&bookm[mid.x][mid.y]!=INT_MAX)        {            aans=min(bookm[mid.x][mid.y]+booky[mid.x][mid.y],aans);            ans=min(aans,ans);        }    }    cout << ans*11 << endl;    }}