HDU - 2612 Find a way 【双向bfs】

题目简述

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.
‘@’ KCF

输出

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.

分析

分别对两个人的位置跑一边bfs，一旦交汇到KFC处就更新一下答案。

与传统双向bfs不同的地方是需要更新最小路径。

AC代码：

点击查看代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
#define N 205
#define inf 1145141919
struct node{
	int x,y,id;
}s,r;
const int dx[]={-1,0,0,1};
const int dy[]={0,-1,1,0};
int dis1[N][N],dis2[N][N];
char ch[N][N];
int ans=inf,n,m;
void bfs(){
	queue<node>q;
	q.push(s);q.push(r);
	while(!q.empty()){
		node temp=q.front();q.pop();
		int tempx=temp.x,tempy=temp.y;
		if(dis1[tempx][tempy]&&dis2[tempx][tempy]&&ch[tempx][tempy]=='@'){
			ans=min(ans,dis1[tempx][tempy]+dis2[tempx][tempy]);
		}
		for(int i=0;i<=3;i++){
			int nx=tempx+dx[i],ny=tempy+dy[i],id=temp.id;
			if(id==1){
				if(nx<=0 || nx>n || ny<=0 || ny>m|| ch[nx][ny]=='#'|| dis1[nx][ny]) continue;
				node curr;curr.id=id;curr.x=nx;curr.y=ny;
				q.push(curr);
				dis1[curr.x][curr.y]=dis1[tempx][tempy]+1;
			}
			else if(id==2){
				if(nx<=0 || nx>n || ny<=0 || ny>m|| ch[nx][ny]=='#'|| dis2[nx][ny]) continue;
				node curr;curr.id=id;curr.x=nx;curr.y=ny;
				q.push(curr);
				dis2[curr.x][curr.y]=dis2[tempx][tempy]+1;
			}
		}
	}
}
int main(){
	while(cin>>n>>m){
		ans=inf;
		memset(dis1,0,sizeof(dis1));
		memset(dis2,0,sizeof(dis2));
		for(int i=1;i<=n;i++)
			for(int j=1;j<=m;j++){
				cin>>ch[i][j];
				if(ch[i][j]=='Y'){
					s.id=1;s.x=i;s.y=j;
				}
				if(ch[i][j]=='M'){
					r.id=2;r.x=i;r.y=j;
				}
			}
		bfs();
		cout<<11*ans<<endl;
	}
	return 0;
}