(简单) HDU 2612 Find a way，BFS。

　　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.

 

　　就是求两个人到某一个KFC的最小值，这个题记得以前做的时候被坑惨了，要注意初始化为INF，以及说人的起始位置要表示为可通过。

 

代码如下：

#include<iostream>
#include<cstring>

using namespace std;

const int INF=1e7;

short map1[205][205];
int que[40005],las,fir;
int ans[205][205];
int ans1[205][205];
int N,M;
int couKFC,Si,Sj,Ei,Ej;

bool judge(int x,int y,int (*rem)[205])
{
    if(x<=0||y<=0||x>N||y>M)
        return 0;

    if(rem[x][y]!=INF)
        return 0;

    if(map1[x][y]==0)
        return 0;

    return 1;
}

void bfs(int x,int y,int (*rem)[205])
{
    las=fir=0;
    int cou=0;
    int temp,t1,t2;

    que[las++]=x*1000+y;
    rem[x][y]=0;

    while(las-fir)
    {
        temp=que[fir++];
        t1=temp/1000;
        t2=temp%1000;
        temp=rem[t1][t2];

        if(map1[t1][t2]==2)
            ++cou;

        if(cou>=couKFC)
            return;

        --t1;
        if(judge(t1,t2,rem))
        {
            rem[t1][t2]=temp+1;
            que[las++]=t1*1000+t2;
        }    
        t1+=2;
        if(judge(t1,t2,rem))
        {
            rem[t1][t2]=temp+1;
            que[las++]=t1*1000+t2;
        }    
        --t1;
        --t2;
        if(judge(t1,t2,rem))
        {
            rem[t1][t2]=temp+1;
            que[las++]=t1*1000+t2;
        }    
        t2+=2;
        if(judge(t1,t2,rem))
        {
            rem[t1][t2]=temp+1;
            que[las++]=t1*1000+t2;
        }
    }
}

int slove()
{
    bfs(Si,Sj,ans);
    bfs(Ei,Ej,ans1);

    int minn=INF;

    for(int i=1;i<=N;++i)
        for(int j=1;j<=M;++j)
            if(map1[i][j]==2)
                if(minn>ans[i][j]+ans1[i][j])
                    minn=ans[i][j]+ans1[i][j];

    return minn*11;
}

int main()
{
    ios::sync_with_stdio(false);

    char c;

    while(cin>>N>>M)
    {
        couKFC=0;

        for(int i=1;i<=N;++i)
            for(int j=1;j<=M;++j)
            {
                cin>>c;
                ans[i][j]=ans1[i][j]=INF;

                switch(c)
                {
                    case 'Y':
                        map1[i][j]=1;
                        Si=i;
                        Sj=j;
                        break;
                    case 'M':
                        map1[i][j]=1;
                        Ei=i;
                        Ej=j;
                        break;
                    case '.':
                        map1[i][j]=1;
                        break;
                    case '#':
                        map1[i][j]=0;
                        break;
                    case '@':
                        map1[i][j]=2;
                        ++couKFC;
                        break;
                }
            }

        cout<<slove()<<endl;
    }

    return 0;
}