Find a way

Time Limit : 3000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 48 Accepted Submission(s) : 15

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

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

———————————————————————————————————————————————————

题目说两个人，他们想在某家KCF见面，要求他们到这家KCF的时间花费最少。

我们从每个kfc出发搜到每个人，选出到两个人距离和最小的就是答案

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
using namespace std;char mp[205][205];
int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int vir[205][205];
int m,n,ans1,ans2,mn;struct node
{int x,y,cnt;
};bool cheak(int i,int j)
{if(i<1||i>m||j<1||j>n||mp[i][j]=='#')return 0;elsereturn 1;
}void bfs(int si,int sj,int yi,int yj,int mi,int mj)
{queue<node>q;node f,d;memset(vir,0,sizeof(vir));f.x=si;f.y=sj;f.cnt=0;vir[f.x][f.y]=1;q.push(f);while(!q.empty()){f=q.front();q.pop();if(ans1==-1&&ans2==-1&&2*f.cnt>=mn)return;if(ans1==-1&&ans2!=-1&&f.cnt+ans2>=mn)return;if(ans1!=-1&&ans2==-1&&f.cnt+ans1>=mn)return;if(f.x==yi&&f.y==yj&&ans1==-1){ans1=f.cnt;}if(f.x==mi&&f.y==mj&&ans2==-1){ans2=f.cnt;}if(ans1!=-1&&ans2!=-1)return;for(int i=0;i<4;i++){d.x=f.x+dir[i][0];d.y=f.y+dir[i][1];if(cheak(d.x,d.y)&&!vir[d.x][d.y]){d.cnt=f.cnt+1;vir[d.x][d.y]=1;q.push(d);}}}return ;
}int main()
{int yi,yj,mi,mj,ans;while(scanf("%d%d",&m,&n)!=EOF){for(int i=1;i<=m;i++)for(int j=1;j<=n;j++){cin>>mp[i][j];if(mp[i][j]=='Y'){yi=i;yj=j;}if(mp[i][j]=='M'){mi=i;mj=j;}}mn=100000;for(int i=1;i<=m;i++)for(int j=1;j<=n;j++){if(mp[i][j]=='@'){ans1=-1;ans2=-1;bfs(i,j,yi,yj,mi,mj);if(ans1!=-1&&ans2!=-1){ans=ans1+ans2;if(ans<mn)mn=ans;}}}printf("%d\n",mn*11);}return 0;
}

Hdu2612 Find a way相关推荐

HDU2612 Find a Way BFS
题意: yifenfei和merceki要去KFC聚会,给出一个地方的地图,n*m,有若干个个KFC,然后他们每走一步需要11分钟(注意:这里时间不能重叠的,比如yi走了一步,me也是走了一步,则一共 ...
HDU-2612 Find a way
Problem Description Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. L ...
HDU2612(BFS算法)
Problem Descrption Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Le ...
hdu2612——Find a way
Problem Description Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. L ...
Find a way hdu2612
Find a way Time Limit : 3000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other) Total Su ...
22.11.27补卡 HDU-2612 bfs
这题dfs没法写, 因为dfs没法处理最优解的情况分别记录每个人走每一个kfc的最短路径然后一个双重循环搜最小值就好了初始化记得初始化为最大值, 不然有可能影响最min取值 ps: 这题我用st ...
【HDU - 2612】Find a way（bfs）
-->Find a way 直接上Chinese Descriptions: hsj和lsh最近迷上了pokemon go的游戏.在双十一大物期中考试来临之前,他们想抓一只稀有土拨鼠来攒攒人品 ...
ACM训练计划建议（转）
ACM训练计划建议 From:freecode# Date:2015/5/20 前言: 老师要我们整理一份训练计划给下一届的学弟学妹们,整理出来了,费了不少笔墨,就也将它放到博客园上供大家参考. 菜 ...
ACM的分类训练题集
1.数论大概有素数测试(筛法),扩展欧几里得算法,同余模运算,高斯消元,中国剩余定理,莫比乌斯反演等等. 我不擅长这方面(数学烂,还好后期团队里有两位数学大神),不发表评论. 推荐题目: 同余模运算 ...

Hdu2612 Find a way