本文主要是介绍hdu 2612 Find a way【BFS】，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2612
题意：y和m要在kfc见面，问你能否选择一家kfc使得y和m到那里的距离最短
解析：两个人都跑一遍bfs，然后枚举每家kfc，找出最优的即可
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
const int inf = 0x7fffffff;
int dx[] = {0,1,-1,0};
int dy[] = {1,0,0,-1};
int vis[205][205];
int step1[205][205];
int step2[205][205];
int n,m;
char a[205][205];
struct node
{int x,y;int step;node() {}node(int _x,int _y,int _step){x = _x;y = _y;step = _step;}
};
void bfs(int sx,int sy,int step[205][205])
{memset(vis,0,sizeof(vis));queue<node>q;q.push(node(sx,sy,0));while(!q.empty()){node now = q.front();q.pop();if(a[now.x][now.y]=='@'){if(step[now.x][now.y])step[now.x][now.y] = min(step[now.x][now.y],now.step);elsestep[now.x][now.y] = now.step;}for(int i=0;i<4;i++){int tx = now.x+dx[i];int ty = now.y+dy[i];if(tx<0 || tx>=n || ty<0 || ty>=m)continue;if(a[tx][ty]=='#' || vis[tx][ty])continue;vis[tx][ty] = 1;q.push(node(tx,ty,now.step+1));}}
}
int main()
{while(~scanf("%d %d",&n,&m)){int x1,y1,x2,y2;for(int i=0;i<n;i++){scanf("%s",a[i]);for(int j=0;j<m;j++){if(a[i][j]=='Y'){x1 = i;y1 = j;}if(a[i][j]=='M'){x2 = i;y2 = j;}}}memset(step1,0,sizeof(step1));memset(step2,0,sizeof(step2));bfs(x1,y1,step1);bfs(x2,y2,step2);int ans = inf;for(int i=0;i<n;i++){for(int j=0;j<m;j++){if(a[i][j] == '@'){if(step1[i][j] && step2[i][j])ans = min(ans,step1[i][j]+step2[i][j]);}}}printf("%d\n",ans*11);}return 0;
}
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