Find a way

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Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2252    Accepted Submission(s): 713

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
   

 

Author

yifenfei

 

Source

奋斗的年代

 

Recommend

yifenfei

此码非本人制作，最近正这研究搜索，这个代码看了大半天看懂了一半多，非常好，值得珍藏，日后好好研究。

#include <iostream>
#include <queue>
#include <string.h>
#include <stdio.h>
using namespace std;
#define N 205

struct node
{
	int x,y,step;
};

char a[N][N];
node sss[1000];
int ans1[1001],ans2[1001],ans[1001],mark[N][N];
int x1,y1,x2,y2,Min,cnt;
int dis[4][2]={{0,1},{1,0},{0,-1},{-1,0}};

int judge(int x,int y)
{
	int i;
	for(i=1;i<cnt;i++)
	{
		if(sss[i].x==x && sss[i].y==y)
			return i;
	}
	return 0;
}
void BFS_Y(int x,int y)
{
	queue<node> q;
	node s1,s2;
	s1.x=x;
	s1.y=y;
	s1.step=0;
	q.push(s1);
	int i;
	memset(ans,0,sizeof(ans));
	memset(mark,0,sizeof(mark));
	mark[s1.x][s1.y]=1;
	while(!q.empty())
	{
		s1=q.front();
		q.pop();
		int t=judge(s1.x,s1.y);
		if(t)
		{
			mark[s1.x][s1.y]=1;
			ans[t]=s1.step;
		}
		for(i=0;i<4;i++)
		{
			s2=s1;
			s2.x+=dis[i][0];
			s2.y+=dis[i][1];
			s2.step++;
			if((a[s2.x][s2.y]=='.' || a[s2.x][s2.y]=='@') && mark[s2.x][s2.y]==0)
			{
				q.push(s2);
				mark[s2.x][s2.y]=1;
			}
		}
	}
	for(i=1;i<cnt;i++)
	{
		if(ans[i]) ans1[i]=ans[i];
		else ans1[i]=-1;
	}
}

void BFS_M(int x,int y)
{
	queue<node> q;
	node s1,s2;
	s1.x=x;
	s1.y=y;
	s1.step=0;
	q.push(s1);
	int i;
	memset(ans,0,sizeof(ans));
	memset(mark,0,sizeof(mark));
	mark[s1.x][s1.y]=1;
	while(!q.empty())
	{
		s1=q.front();
		q.pop();
		int t=judge(s1.x,s1.y);
		if(t)
		{
			mark[s1.x][s1.y]=1;
			ans[t]=s1.step;
		}
		for(i=0;i<4;i++)
		{
			s2=s1;
			s2.x+=dis[i][0];
			s2.y+=dis[i][1];
			s2.step++;
			if((a[s2.x][s2.y]=='.' || a[s2.x][s2.y]=='@') && mark[s2.x][s2.y]==0)
			{
				q.push(s2);
				mark[s2.x][s2.y]=1;
			}
		}
	}
	for(i=1;i<cnt;i++)
	{
		if(ans[i]) ans2[i]=ans[i];
		else ans2[i]=-1;
	}
}

int main()
{

	int n,m,i,j;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		getchar();
		Min=99999;
		cnt=1;
		memset(a,'#',sizeof(a));
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=m;j++)
			{
				scanf("%c",&a[i][j]);
				if(a[i][j]=='Y')
				{
					x1=i;y1=j;
				}
				else if(a[i][j]=='M')
				{
					x2=i;y2=j;
				}
				else if(a[i][j]=='@')
				{
					sss[cnt].x=i;
					sss[cnt++].y=j;
				}
			}
			getchar();
		}
		BFS_Y(x1,y1);
		BFS_M(x2,y2);
		for(i=1;i<cnt;i++)
		{
			if(ans1[i]!=-1 && ans2[i]!=-1)
			{
				int t=ans1[i]+ans2[i];
				if(t<Min)
					Min=t;
			}
		}
		printf("%d\n",Min*11);
	}
	return 0;
}