Find a way

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 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#

Sample Output

66
88
66

这个题最开始的想法就是对分别对各个kfc的位置进行bfs，意料之中的TLE了，然而就看了一个博客，对每个位置都标记步数，然后对各个@位置求总步数的最小值，这个最小值如果初始化为0时，要把步数为0 的结果值去掉，原因1步数不可能为0 原因2为0要不表示没有路要不表示到不了这个地方，所以要去除为0的情矿

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
#include <cmath>
#define maxx 210
using namespace std;

int n,m;
int book_y[maxx][maxx],book_m[maxx][maxx];
char ma[maxx][maxx];

struct note
{
    int x,y,step;
}st,ed;

void bfs_y()
{
    int i;
    int next[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
    queue<note>Q;
    Q.push(st);
    while(!Q.empty())
    {
        st=Q.front();
        Q.pop();
        for(i=0;i<4;i++)
        {
            ed.x=st.x+next[i][0];
            ed.y=st.y+next[i][1];
            if(book_y[ed.x][ed.y]==0&&(ma[ed.x][ed.y]=='.'||ma[ed.x][ed.y]=='@')&&ed.x>=0&&ed.y>=0&&ed.x<n&&ed.y<m)
            {
                ed.step=st.step+1;
                Q.push(ed);
                book_y[ed.x][ed.y]=ed.step;;
            }
        }
    }
}

void bfs_m()
{
    int i;
    int next[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
    queue<note>Q;
    Q.push(st);
    while(!Q.empty())
    {
        st=Q.front();
        Q.pop();
        for(i=0;i<4;i++)
        {
            ed.x=st.x+next[i][0];
            ed.y=st.y+next[i][1];
            if(book_m[ed.x][ed.y]==0&&ma[ed.x][ed.y]!='#'&&ed.x>=0&&ed.y>=0&&ed.x<n&&ed.y<m)
            {
                ed.step=st.step+1;
                Q.push(ed);
                book_m[ed.x][ed.y]=ed.step;;
            }
        }
    }
}

int main()
{
    char c;
    int i,j,yy,yx,mx,my;
    while(~scanf("%d %d",&n,&m))
    {
        memset(ma,0,sizeof(ma));
        for(i=0;i<n;i++)
        {
            getchar();
            for(j=0;j<m;j++)

            {
                scanf("%c",&c);
                ma[i][j]=c;
                if(c=='Y')
                {
                    yx=i;yy=j;
                }
                else if(c=='M')
                {
                    mx=i;my=j;
                }
            }
        }
        memset(book_y,0,sizeof(book_y));
        memset(book_m,0,sizeof(book_m));
        st.x=yx,st.y=yy,st.step=0;book_y[yx][yy]=1;
        bfs_y();
        st.x=mx,st.y=my,st.step=0;book_m[mx][my]=1;
        bfs_m();
        int ans=0x3f3f3f3f;
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(ma[i][j]=='@'&&book_y[i][j]!=0&&book_m[i][j]!=0)
                {
                    ans=min(ans,book_y[i][j]+book_m[i][j]);
                }
            }
        }
        printf("%d\n",ans*11);
    }
    return 0;
}

这个代码再深入思考一下就应该考虑一个八两个代码合成一个代码，本来想着数组不一样不能合成，再翻了几个博客，发现直接步数相加就可以了

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
#include <cmath>
#define maxx 210
#define inf 0x3f3f3f3f
using namespace std;

int n,m;
int visit[maxx][maxx],book[maxx][maxx];
char ma[maxx][maxx];

struct note
{
    int x,y,step;
}st,ed;

void bfs(int num)
{
    int i;
    int next[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
    memset(visit,0,sizeof(visit));
    visit[st.x][st.y]=1;
    queue<note>Q;
    Q.push(st);
    while(!Q.empty())
    {
        st=Q.front();
        Q.pop();
        if(ma[st.x][st.y]=='@')
        {
            if(num==1)
            {
                book[st.x][st.y]=st.step;
            }
            else if(book[st.x][st.y]!=-1)
            {
                book[st.x][st.y]+=st.step;
            }
        }
        for(i=0;i<4;i++)
        {
            ed.x=st.x+next[i][0];
            ed.y=st.y+next[i][1];
            if(visit[ed.x][ed.y]==0&&ma[ed.x][ed.y]!='#'&&ed.x>=0&&ed.y>=0&&ed.x<n&&ed.y<m)
            {
                ed.step=st.step+1;
                Q.push(ed);
                visit[ed.x][ed.y]=1;
            }
        }
    }
}

int main()
{
    char c;
    int i,j,yy,yx,mx,my;
    while(~scanf("%d %d",&n,&m))
    {
        memset(ma,0,sizeof(ma));
        for(i=0;i<n;i++)
        {
            getchar();
            for(j=0;j<m;j++)

            {
                scanf("%c",&c);
                ma[i][j]=c;
                if(c=='Y')
                {
                    yx=i;yy=j;
                }
                else if(c=='M')
                {
                    mx=i;my=j;
                }
            }
        }
        memset(book,-1,sizeof(book));
        st.x=yx,st.y=yy,st.step=0;
        bfs(1);
        st.x=mx,st.y=my,st.step=0;
        bfs(2);
        int ans=inf;
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(ma[i][j]=='@'&&book[i][j]!=-1&&visit[i][j]==1)//第一个标注Y一定路过，第二个表示第二个一定路过
                {
                    ans=min(ans,book[i][j]);
                }
            }
        }
        printf("%d\n",ans*11);
    }
    return 0;
}