Find a way（bfs）

傷城~ 2022-08-02 09:49 44阅读 0赞

# Find a way #

**Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 5710    Accepted Submission(s): 1905**

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

Sample Output

66
    88
    66

Author

yifenfe

这一题很巧妙地用到了三维数组用来村每一个@M Y到这个点的时间，还有一个问题没有解决！

#include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<queue>
    const int infinity=0x3f3f3f3f;
    const int maxs=210;
    using namespace std;
    int dir[4][2]={0,1,0,-1,1,0,-1,0};
    int n,m,x1,y1,x2,y2;
    char map[maxs][maxs];
    int vis[maxs][maxs];
    int num[maxs][maxs][2];
    struct node
    {
        int x;
        int y;
        int step;
    };
    int check(int x,int y)
    {
        if(x<0||x>=n||y<0||y>=m)
        {
            return 1;
        }
        if(vis[x][y]||map[x][y]=='#')
        {
            return 1;
        }
        return 0;
    }
    void bfs(int x1,int y1,int flag)
    {
        int i;
        queue<node>Q;
        node rencent,next;
        rencent.x=x1;
        rencent.y=y1;
        rencent.step=0;
        vis[rencent.x][rencent.y]=1;
        Q.push(rencent);
        while(!Q.empty())
        {
           // cout<<"ok"<<endl;
            rencent=Q.front();
            Q.pop();
            if(map[rencent.x][rencent.y]=='@')
            {
               // cout<<rencent.step<<endl;
                num[rencent.x][rencent.y][flag]=rencent.step;
            }
            for(i=0;i<4;i++)
            {
                next=rencent;
                next.x+=dir[i][0];
                next.y+=dir[i][1];
                if(check(next.x,next.y))
                {
                    continue;
                }
                next.step=rencent.step+1;
                vis[next.x][next.y]=1;
                Q.push(next);
            }
        }
    
    }
    int main()
    {
        int i,j,k,mins;
        while(cin>>n>>m)
        {
            mins=infinity;
            for(i=0;i<n;i++)
            {
                for(j=0;j<m;j++)
                {
                    cin>>map[i][j];
                    if(map[i][j]=='Y')
                    {
                        x1=i;y1=j;
                    }
                    if(map[i][j]=='M')
                    {
                        x2=i;y2=j;
                    }
                }
            }
            for(i=0;i<n;i++)
            {
                for(j=0;j<m;j++)
                {
                    for(k=0;k<2;k++)
                    {
                        num[i][j][k]=0;
                    }
                }
            }
            memset(vis,0,sizeof(vis));
            bfs(x1,y1,0);
            memset(vis,0,sizeof(vis));
            bfs(x2,y2,1);
            for(i=0;i<n;i++)
            {
                for(j=0;j<m;j++)
                {
                    if(map[i][j]=='@'&&num[i][j][0]+num[i][j][1]<mins)
                    {
                        mins=num[i][j][0]+num[i][j][1];
                    }
                }
            }
            cout<<mins*11<<endl;
        }
    }