Find a way

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24190 Accepted Submission(s): 7911

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

Sample Output

Author

yifenfei

Source

奋斗的年代

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Statistic | Submit | Discuss | Note

我认为，M 能及经过 Y ，Y 也能到 M

用BFS寻找 $\ Y\ or\ M$ 到每一个 $@$ 的最短距离
然后记录 $\ Y\$ 能到达每一个 $\ @\$ 的距离，用 $\ cnt1[i][j]\$ 来记录，其中， $i,j$ ，是当前 $@$ 的坐标

然后记录 $\ M\$ 能到达每一个 $\ @\$ 的距离，用 $\ cnt2[i][j]\$ 来记录，其中， $i,j$ ，是当前 $@$ 的坐标

找到之后遍历整张图

ans=0x3f3f3f3f;
for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
                if(str[i][j]=='@')
                    ans=min(ans,cnt1[i][j]+cnt2[i][j]);

CODE：

#include<bits/stdc++.h>
using namespace std;
const int MAXN=222;
char str[MAXN][MAXN];
bool vis[MAXN][MAXN];
int cnt1[MAXN][MAXN];
int cnt2[MAXN][MAXN];

int d[4][2]={1,0, -1,0, 0,1, 0,-1};
int n,m,sx,sy,yx,yy,mx,my;

struct node{
    int x,y,step;
    node(){};
    node(int _x,int _y,int _step)
    {
        x=_x; y=_y; step=_step;
    }
    bool operator <(const node &b)const
    {
        return step>b.step;
    }
};

void  bfs1(int x,int y)
{
    memset(vis,0,sizeof(vis));
    memset(cnt1,0x3f,sizeof(cnt1));//需要对其初始化为 INF 因为有可能一个能到，另外一个不能到
    vis[x][y]=1;
    priority_queue<node> que;
    que.push(node(x,y,0));
    node e1,e2;
    int ans=0;
    while(que.size())
    {
        e1=que.top();
        que.pop();
        if(str[e1.x][e1.y]=='@')
        {
            vis[e1.x][e1.y]=1;
            cnt1[e1.x][e1.y]=e1.step;//记录 Y 到 @ 的最小距离
        }
        for(int i=0;i<4;i++)
        {
            e2.x=e1.x+d[i][0];
            e2.y=e1.y+d[i][1];
            e2.step=e1.step+1;
            if(!vis[e2.x][e2.y] && str[e2.x][e2.y]!='#' && 0 <= e2.x && e2.x <n && 0 <= e2.y && e2.y <m)
            {
                vis[e2.x][e2.y]=1;
                que.push(e2);
            }
        }
    }
}
void  bfs2(int x,int y)
{
    memset(vis,0,sizeof(vis));
    memset(cnt2,0x3f,sizeof(cnt2));
    vis[x][y]=1;
    priority_queue<node> que;
    que.push(node(x,y,0));
    node e1,e2;
    int ans=0;
    while(que.size())
    {
        e1=que.top();
        que.pop();
        if(str[e1.x][e1.y]=='@')
        {
            vis[e1.x][e1.y]=1;
            cnt2[e1.x][e1.y]=e1.step;//记录 M 到 @ 的最小距离
        }
        for(int i=0;i<4;i++)
        {
            e2.x=e1.x+d[i][0];
            e2.y=e1.y+d[i][1];
            e2.step=e1.step+1;
            if(!vis[e2.x][e2.y] && str[e2.x][e2.y]!='#' && 0 <= e2.x && e2.x <n && 0 <= e2.y && e2.y <m)
            {
                vis[e2.x][e2.y]=1;
                que.push(e2);
            }
        }
    }
}
int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        for(int i=0;i<n;i++)    scanf("%s",str[i]);
//        for(int i=0;i<n;i++)    printf("%s\n",str[i]);
        int ans=0x3f3f3f3f;
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
            {
                if(str[i][j]=='Y')
                    bfs1(i,j);
                if(str[i][j]=='M')
                    bfs2(i,j);
            }
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
                if(str[i][j]=='@')
                    ans=min(ans,cnt1[i][j]+cnt2[i][j]);
        printf("%d\n",ans*11);

    }
    return 0;
}

HDU 2612 Find a way ——————BFS + 打表

Find a way

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