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2014
02-12

hdu 2612 Find a way-BFS-[解题报告]C++

Find a way

问题描述 :

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.

输入:

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.
[email protected] KCF

输出:

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.
[email protected] KCF

样例输入:

4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#

样例输出:

66
88
66

/*
分析:
    这得犯多大的2,才能,在前面都处理成n*m,后面
就处理成了n*n,还WA3次后才发现啊。。。

    进行两次广搜,取最小和
                                       2012-07-19
*/

#include"stdio.h"
#include"queue"
#include"string.h"
using namespace std;
int dir[4][2]={1,0, -1,0, 0,1, 0,-1};


int ori[211][211];
int map[211][211];
int x_s1,y_s1;
int x_s2,y_s2;
int n,m;


int ans[211][211];


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


int min(int a,int b)
{
	return a>b?b:a;
}
int judge(int x,int y)
{
	if(x<0 || x>=n || y<0 || y>=m)	return 1;
	if(map[x][y]==1)	return 1;
	return 0;
}


void BFS(int x,int y)
{
	queue<node>q;
	node cur,next;
	int i;


	cur.x=x;
	cur.y=y;
	cur.step=0;


	map[cur.x][cur.y]=1;
	q.push(cur);
	while(!q.empty())
	{
		cur=q.front();
		q.pop();
		if(ori[cur.x][cur.y]==2)	ans[cur.x][cur.y]+=cur.step;
		for(i=0;i<4;i++)
		{
			next.x=cur.x+dir[i][0];
			next.y=cur.y+dir[i][1];


			if(judge(next.x,next.y))	continue;
			next.step=cur.step+1;
			map[next.x][next.y]=1;
			q.push(next);
		}
	}
}


int main()
{
	int i,l;
	char str[211];
	int Ans;


	while(scanf("%d%d",&n,&m)!=-1)
	{
		memset(ori,0,sizeof(ori));
		memset(ans,0,sizeof(ans));
		for(i=0;i<n;i++)
		{
			scanf("%s",str);
			for(l=0;str[l];l++)
			{
				if(str[l]=='Y')		{x_s1=i;y_s1=l;}
				else if(str[l]=='M'){x_s2=i;y_s2=l;}
				else if(str[l]=='.')ori[i][l]=0;
				else if(str[l]=='#')ori[i][l]=1;
				else if(str[l]=='@')ori[i][l]=2;
			}
		}


		for(i=0;i<n;i++)
		for(l=0;l<m;l++)
			map[i][l]=ori[i][l];
		BFS(x_s1,y_s1);


		for(i=0;i<n;i++)
		for(l=0;l<m;l++)
			map[i][l]=ori[i][l];
		BFS(x_s2,y_s2);


		Ans=11111111;
		for(i=0;i<n;i++)
		{
			for(l=0;l<m;l++)
			{
				if(ans[i][l])
				{
					Ans=min(Ans,ans[i][l]);
				}
			}
		}
		printf("%d\n",Ans*11);
	}
	return 0;
}

解题转自:http://blog.csdn.net/ice_crazy/article/details/7762700


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  1. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮