Find a way（两路广搜）

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

上述每人只需要搜一次并记录下到每一个@的步数和，最后一比较就好了！

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=202;
const int INF=0x3f3f3f3f;
char map1[maxn][maxn],vis[maxn][maxn];
int dis[maxn][maxn][2];                   //记录下来每个@所走的步数，其中第三位代表的是哪个人在走
int n,m,flag;
struct node
{
    int x,y,step;
};
bool judge(int x,int y)
{
    if(x>=0&&x<n&&y>=0&&y<m&&!vis[x][y]&&map1[x][y]!='#')
        return 1;
    return 0;
}
int dxy[][2]= {0,1,0,-1,1,0,-1,0};
void bfs(int x,int y)
{
    node now,next;
    queue<node>q;
    now.x=x,now.y=y,now.step=0;
    q.push(now);
    vis[x][y]=1;
    while(!q.empty())
    {
        now=q.front();
        if(map1[now.x][now.y]=='@')                 //遇到@就记录下来走了多少步
            dis[now.x][now.y][flag]=now.step;
        for(int i=0; i<4; i++)
        {
            int xx=now.x+dxy[i][0],yy=now.y+dxy[i][1];
            if(judge(xx,yy))
            {
                vis[xx][yy]=1;
                next.x=xx,next.y=yy,next.step=now.step+1;
                q.push(next);
            }
        }
        q.pop();
    }
}
int main()
{
	ios::sync_with_stdio(false);
    while(cin>>n>>m)
    {
        int i,j,s=INF;
        for(i=0; i<n; i++)
            for(j=0; j<m; j++)
                dis[i][j][0]=dis[i][j][1]=INF;  //有的@可能是无法到达的，所以用无穷大初始化！
        for(i=0; i<n; i++)
           cin>>map1[i];
        for(i=0; i<n; i++)
            for(j=0; j<m; j++)
            {
                if(map1[i][j]=='Y')
                {
                    flag=0;
                    memset(vis,0,sizeof(vis));
                    bfs(i,j);
                }
                if(map1[i][j]=='M')
                {
                    flag=1;
                    memset(vis,0,sizeof(vis));
                    bfs(i,j);
                }
            }
        for(i=0; i<n; i++)
            for(j=0; j<m; j++)
            {
                if(map1[i][j]=='@'&&s>dis[i][j][0]+dis[i][j][1])    //求最小的步数，相比较
                    s=dis[i][j][0]+dis[i][j][1];
            }
       cout<<s*11<<endl;
    }
    return 0;
}