HDU：2612 Find a way（双BFS+打表）

Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10413    Accepted Submission(s): 3413

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

题目大意：y和m要去肯德基聚餐，图中有多个kfc，他们要选的那个kfc必须到彼此的所用时间之和最小，问  最少需要多少时间（这里一格代表了11分钟）。

解题思路：将y到每点的时间和m到每点的时间用bfs打表，打好表后扫地图，如果扫到@的话算下时间之和，更新min。

（ 这题有点坑，y不能经过m的出发点，m也不能经过y的出发点，详情看http://blog.youkuaiyun.com/wjw0130/article/details/36254119）

代码如下：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
char map[210][210];
int visit[210][210];
int cnt1[210][210];//储存y到每点的步数 
int cnt2[210][210];//储存m到每点的步数 
int n,m;
int cnt,yn,ym,mn,mm;
struct node
{
	int n,m;
	int step;
};
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		cnt=0;
		int ans=INF;
		memset(cnt1,INF,sizeof(cnt1));
		memset(cnt2,INF,sizeof(cnt2));
		for(int i=0;i<n;i++)
		{
			scanf("%s",&map[i]);
			for(int j=0;j<m;j++)
			{
				if(map[i][j]=='Y')
				{
					yn=i;
					ym=j;
				}
				if(map[i][j]=='M')
				{
					mn=i;
					mm=j;
				}
			}
		}
			memset(visit,0,sizeof(visit));
			struct node start;
			start.n=yn;
			start.m=ym;
			start.step=0;
			visit[yn][ym]=1;
			map[mn][mm]='#';//这里，注意下，y走路的时候不能经过m的路，所以临时把m的坐标变成# 
			queue<node>qy;
			qy.push(start);
			while(!qy.empty())//开始给y打表 
			{
				struct node tmp=qy.front();
				qy.pop();
				cnt1[tmp.n][tmp.m]=tmp.step;
				tmp.step++;
				struct node tmp2=tmp;
				if(tmp2.n-1>=0&&visit[tmp2.n-1][tmp2.m]==0&&map[tmp2.n-1][tmp2.m]!='#')
				{
					visit[tmp2.n-1][tmp2.m]=1;
					cnt1[tmp2.n-1][tmp2.m]=tmp2.step;
					tmp2.n--;
					qy.push(tmp2);
				}
				tmp2=tmp;
				if(tmp2.n+1<n&&visit[tmp2.n+1][tmp2.m]==0&&map[tmp2.n+1][tmp2.m]!='#')
				{
					visit[tmp2.n+1][tmp2.m]=1;
					cnt1[tmp2.n+1][tmp2.m]=tmp2.step;
					tmp2.n++;
					qy.push(tmp2);
				}
				tmp2=tmp;
				if(tmp2.m-1>=0&&visit[tmp2.n][tmp2.m-1]==0&&map[tmp2.n][tmp2.m-1]!='#')
				{
					visit[tmp2.n][tmp2.m-1]=1;
					cnt1[tmp2.n][tmp2.m-1]=tmp2.step;
					tmp2.m--;
					qy.push(tmp2);
				}
				tmp2=tmp;
				if(tmp2.m+1<m&&visit[tmp2.n][tmp2.m+1]==0&&map[tmp2.n][tmp2.m+1]!='#')
				{
					visit[tmp2.n][tmp2.m+1]=1;
					cnt1[tmp2.n][tmp2.m+1]=tmp2.step;
					tmp2.m++;
					qy.push(tmp2);
				}
			}
			map[mn][mm]='M';//还原m 
			memset(visit,0,sizeof(visit));//这里记得在清0以下 
			start.n=mn;
			start.m=mm;
			start.step=0;
			visit[mn][mm]=1;
			map[yn][ym]='#';//m不能走y所在的地址 
			queue<node>qm;
			qm.push(start);
			while(!qm.empty())//开始给m打表 
			{
				struct node tmp=qm.front();
				qm.pop();
				tmp.step++;
				struct node tmp2=tmp;
				if(tmp2.n-1>=0&&visit[tmp2.n-1][tmp2.m]==0&&map[tmp2.n-1][tmp2.m]!='#')
				{
					visit[tmp2.n-1][tmp2.m]=1;
					cnt2[tmp2.n-1][tmp2.m]=tmp2.step;
					tmp2.n--;
					qm.push(tmp2);
				}
				tmp2=tmp;
				if(tmp2.n+1<n&&visit[tmp2.n+1][tmp2.m]==0&&map[tmp2.n+1][tmp2.m]!='#')
				{
					visit[tmp2.n+1][tmp2.m]=1;
					cnt2[tmp2.n+1][tmp2.m]=tmp2.step;
					tmp2.n++;
					qm.push(tmp2);
				}
				tmp2=tmp;
				if(tmp2.m-1>=0&&visit[tmp2.n][tmp2.m-1]==0&&map[tmp2.n][tmp2.m-1]!='#')
				{
					visit[tmp2.n][tmp2.m-1]=1;
					cnt2[tmp2.n][tmp2.m-1]=tmp2.step;
					tmp2.m--;
					qm.push(tmp2);
				}
				tmp2=tmp;
				if(tmp2.m+1<m&&visit[tmp2.n][tmp2.m+1]==0&&map[tmp2.n][tmp2.m+1]!='#')
				{
					visit[tmp2.n][tmp2.m+1]=1;
					cnt2[tmp2.n][tmp2.m+1]=tmp2.step;
					tmp2.m++;
					qm.push(tmp2);
				}
			}
			for(int i=0;i<n;i++)
			{
				for(int j=0;j<m;j++)
				{
					if(map[i][j]=='@')
					{
						ans=min(ans,cnt1[i][j]+cnt2[i][j]);//更新ans 
					}
				}
			}
			printf("%d\n",ans*11);//一步当11步，题上说的 
	}
	return 0;
}