Find a way

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23967 Accepted Submission(s): 7823

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

题意：两个人去同一个kfc，问两个人一共走的最短路的合，kfc有多个

题解：每个人都要一次bfs，用一个kfc去找人会超时，先把所有的kfc都找到再处理

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<cstring>

#include<sstream>

#include<cmath>

#include<cstdlib>

#include<queue>

#include<map>

#include<set>

using namespace std;

#define INF 0x3f3f3f3f

const int maxn=;

typedef pair<int,int >P;

int n,m;

int xa,xb,ya,yb;

char a[][];

int vis[][];

int d[][];

int d1[][];

struct fwe

{

    int x,y;

}list[];

int dx[]={,,-,};

int dy[]={,,,-};

void bfs(int x,int y)

{

    for(int i=;i<n;i++)

        for(int j=;j<m;j++)

            d[i][j]=INF;

    queue<P>que;

    que.push(P(x,y));

    d[x][y]=;

    while(que.size())

    {

        P p=que.front();

        que.pop();

        for(int i=;i<;i++)

        {

            int nx=p.first+dx[i];

            int ny=p.second+dy[i];

            if(<=nx && nx<n && <=ny && ny<m && a[nx][ny]!='#' && d[nx][ny]==INF)

            {

                que.push(P(nx,ny));

                d[nx][ny]=d[p.first][p.second]+;

            }

        }

    }

}

void bfss(int x,int y)

{

    for(int i=;i<n;i++)

        for(int j=;j<m;j++)

            d1[i][j]=INF;

    queue<P>que;

    for(int i=;i<n;i++)

        for(int j=;j<m;j++)

            d1[i][j]=INF;

    que.push(P(x,y));

    d1[x][y]=;

    while(que.size())

    {

        P p=que.front();

        que.pop();

        for(int i=;i<;i++)

        {

            int nx=p.first+dx[i];

            int ny=p.second+dy[i];

            if(<=nx && nx<n && <=ny && ny<m && a[nx][ny]!='#' && d1[nx][ny]==INF)

            {

                que.push(P(nx,ny));

                d1[nx][ny]=d1[p.first][p.second]+;

            }

        }

    }

}

int main()

{

    int min;

    while(scanf("%d %d",&n,&m)!=EOF)

    {

        for(int i=;i<n;i++)

            for(int j=;j<m;j++)

            {

                cin>>a[i][j];

                if(a[i][j]=='Y')

                {

                    xa=i;

                    ya=j;

                }

                if(a[i][j]=='M')

                {

                    xb=i;

                    yb=j;

                }

            }

        bfs(xa,ya);

        bfss(xb,yb);

        int minn=INF;

        for(int i=;i<n;i++)

            for(int j=;j<m;j++)

            {

                if(a[i][j]=='@')

                {

                    minn=std::min(minn,d[i][j]+d1[i][j]);

                }

            }

        cout<<minn*<<endl;

    }

    return ;

}

巴特西

Find a way HDU - 2612（bfs）

Find a way

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