hdu 2612(Find a way)(简单广搜)

最新推荐文章于 2022-04-12 12:19:28 发布

转载最新推荐文章于 2022-04-12 12:19:28 发布 · 50 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/stepping/p/5669087.html

文章标签：

#java

本文介绍了一种迷宫寻径算法的应用案例，通过双向BFS算法寻找从两个不同起点到多个终点的最短总时间。该算法适用于解决地图中两点到特定目标地点的最短路径问题。

Find a way

Time Limit : 3000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 30 Accepted Submission(s) : 14

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

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

#include <iostream>
#include<cstring>
#include<cstdio>
#include<deque>
#include<algorithm>
using namespace std;
int dr[4][2]={{1,0},{0,1},{-1,0},{0,-1} };
int n,m,i,j,num,x1,x2,y1,y2,ans;
int vis[202][202],a[202][202];
char mp[202][202];
struct node
{
   int x,y,ti;
};
deque<node> s;
void bfs(int x,int y)
{
    int t=num;
    node p;
    p.x=x;
    p.y=y;
    p.ti=0;
    s.clear();
    s.push_back(p);
    vis[x][y]=1;
    while(!s.empty())
    {
        node q=s.front();
        for(i=0;i<4;i++)
        {
            int xx=q.x+dr[i][0];
            int yy=q.y+dr[i][1];
            if(xx>=0 && xx<n && yy>=0 && yy<m && mp[xx][yy]!='#')
                if (!vis[xx][yy])
            {
                p.x=xx;
                p.y=yy;
                p.ti=q.ti+1;
                s.push_back(p);
                vis[xx][yy]=1;
                if(mp[xx][yy]=='@')
                {
                  a[xx][yy]+=p.ti;
                  ans=min(ans,a[xx][yy]);
                  t--;
                }
                if(t==0) return;
            }
        }
        s.pop_front();
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(i=0;i<n;i++)
         {
             scanf("%s",&mp[i]);
             for(j=0;j<m;j++)
              if (mp[i][j]=='Y') x1=i,y1=j;
              else if(mp[i][j]=='M') x2=i,y2=j;
               else if (mp[i][j]=='@') {num++; a[i][j]=0;}
         }
       memset(vis,0,sizeof(vis));
       bfs(x1,y1);
       ans=1000000;
       memset(vis,0,sizeof(vis));
       bfs(x2,y2);
       printf("%d\n",ans*11);
    }
    return 0;
}