HDU 2612 两次BFS

最新推荐文章于 2021-05-26 23:25:51 发布

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最新推荐文章于 2021-05-26 23:25:51 发布

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本文介绍了一个关于两个人从不同起点出发寻找最近的会面地点的问题。通过使用BFS算法两次遍历地图，找到两人到达同一指定地点所需的最短时间。

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Find a way

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2710 Accepted Submission(s): 874

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

Author

yifenfei

题意：Y和M要到@约会，.代表路，#代表墙，求他们到达同一@所需要的最少时间。

进行两次BFS，求出最小的t。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int dir[4][2]={{-1,0},{0,-1},{1,0},{0,1}};
char map[1007][1007];
int vis[1007][1007],ans[1007],ans1[1007],ans2[1007];
int n,m,xy,yy,xm,ym,cnt;
struct sa
{
    int x,y,step;
}s[1007];

int judge(int x,int y)//判断是否到达@ 
{
	for(int i=1;i<cnt;i++)
	{
		if(s[i].x==x&&s[i].y==y)
		return i;
	}
	return 0;
}

void bfs_1(int x,int y)//对Y进行BFS 
{
	queue<sa>q;
	sa s1,s2;
	s1.x=x;
	s1.y=y;
	s1.step=0;
	q.push(s1);
	memset(vis,0,sizeof(vis));
	memset(ans,0,sizeof(ans));
	vis[s1.x][s1.y]=1;
	while(!q.empty())
	{
		s1=q.front();
		q.pop();
		int t=judge(s1.x,s1.y);
		if(t)
		{
			vis[s1.x][s1.y]=1;
			ans[t]=s1.step;
		}
		for(int i=0;i<4;i++)
		{
			s2=s1;
			s2.x+=dir[i][0];
			s2.y+=dir[i][1];
			s2.step++;
			if((map[s2.x][s2.y]=='.'||map[s2.x][s2.y]=='@')&&vis[s2.x][s2.y]==0)
			{
				q.push(s2);
				vis[s2.x][s2.y]=1;
			}
		}
	}
	for(int i=1;i<cnt;i++)
	{
		if(ans[i])ans1[i]=ans[i];
		else ans1[i]=-1;
	}
}

void bfs_2(int x,int y)//对M进行BFS 
{
	queue<sa>q;
	sa s1,s2;
	s1.x=x;
	s1.y=y;
	s1.step=0;
	q.push(s1);
	memset(vis,0,sizeof(vis));
	memset(ans,0,sizeof(ans));
	vis[s1.x][s1.y]=1;
	while(!q.empty())
	{
		s1=q.front();
		q.pop();
		int t=judge(s1.x,s1.y);
		if(t)
		{
			vis[s1.x][s1.y]=1;
			ans[t]=s1.step;
		}
		for(int i=0;i<4;i++)
		{
			s2=s1;
			s2.x+=dir[i][0];
			s2.y+=dir[i][1];
			s2.step++;
			if((map[s2.x][s2.y]=='.'||map[s2.x][s2.y]=='@')&&vis[s2.x][s2.y]==0)
			{
				q.push(s2);
				vis[s2.x][s2.y]=1;
			}
		}
	}
	for(int i=1;i<cnt;i++)
	{
		if(ans[i])ans2[i]=ans[i];
		else ans2[i]=-1;
	}
}
int main()
{
    while (scanf("%d%d",&n,&m)!=EOF)
    {
    	getchar();
		cnt=1;
		for (int i=1;i<=n;i++)
        {
            for (int j=1;j<=m;j++)
            {
                scanf("%c",&map[i][j]);
                if (map[i][j]=='Y')
                {
                    xy=i;
                    yy=j;
                }
                else if (map[i][j]=='M')
                {
                    xm=i;
                    ym=j;
                }
                else if (map[i][j]=='@')
                {
                    s[cnt].x=i;
                    s[cnt++].y=j;
                }
            }
            getchar();
        }
		bfs_1(xy,yy);
		bfs_2(xm,ym);
		int min=99999;
		for(int i=1;i<cnt;i++)//找最小的时间 
		{
			if(ans1[i]!=-1&&ans2[i]!=-1)
			{
				int t=ans1[i]+ans2[i];
				if(min>t)
				min=t;
			}
		}
		printf("%d\n",min*11);
    }
    return 0;
}