HDU 2612 Find a way（[kuangbin带你飞]专题一 简单搜索）

题目连接：题目

题目大意：两个人Y，M去KFC，求两个人到一个KFC的时间和最少；

解题思路：用BFS分别求两个人到每个KFC的距离，然后求最小值

#include <string>
#include<cstdio>
#include <iostream>
#include <deque>
#include<cstring>
using namespace std;
#define size 201

/*节点信息*/
struct node
{
	int x,y,count;
};

char s[size][size];
int dir[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};
int m,n;
int Yx,Yy,Mx,My;/*两个人的坐标*/
int dist1[size][size],dist2[size][size];/*两个人到任意一点的距离*/
bool visited[size][size];
deque<node>Q;


void BFS(node p,int dist[][size])
{
	int i;
	node q;
	memset(visited,0,sizeof(visited));
	memset(dist,0,sizeof(dist));
	Q.push_front(p);
	while (!Q.empty())
	{
		p = Q.front();
		Q.pop_front();
		/*四个方向搜索入队*/
		for (i=0;i<4;i++)
		{
			q.x = p.x+dir[i][0];
			q.y = p.y+dir[i][1];
			q.count = p.count+1;
			if (q.x>=0&&q.y>=0&&q.x<m&&q.y<n&&s[q.x][q.y]!='#'&&!visited[q.x][q.y])
			{
				visited[q.x][q.y] = 1;
				dist[q.x][q.y] = q.count;
				Q.push_back(q);
			}
		}
	}
}

int main()
{
	int i,j,min;
	node p;
	/*freopen("d://1.txt","r",stdin);*/
	while (scanf("%d%d",&m,&n)!=EOF)
	{
		for (i=0;i<m;i++)
		{
			for (j=0;j<n;j++)
			{
				cin>>s[i][j];
				if(s[i][j]=='Y')
					Yx=i,Yy=j;
				else if(s[i][j]=='M')
					Mx = i,My = j;
			}
		}

		min = 999999999;
		p.x = Yx,p.y=Yy,p.count=0;
		BFS(p,dist1);/*求Yifenfei到其他各个点的距离*/
		p.x = Mx,p.y=My,p.count=0;
		BFS(p,dist2);/*求Merceki到其他各个点的距离*/

		for (i=0;i<m;i++)
			for (j=0;j<n;j++)
				if (s[i][j]=='@')/*是KFC，开始计算距离总和*/
					if (dist1[i][j]!=0&&dist2[i][j]!=0)
						if(dist1[i][j]+dist2[i][j]<min)/*比较求距离的最小值*/
							min = dist1[i][j]+dist2[i][j];

		printf("%d\n",min*11);
	}
	return 0;
}