1. 程式人生 > >hdu Find a way(兩路廣搜)

hdu Find a way(兩路廣搜)

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
[email protected]
.#... .#... @..M. #...#
Sample Output 66 88 66 ps:比較笨的我想的是每遇到一次@,兩人就搜一次,結果果斷超時了。。。。 程式碼:

相關推薦

hdu Find a way

Problem Description Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to m

Find a way

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

Find a way個BFS

city any queue ret find 初始化 typedef 維數 opened Problem Description Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo a

1070A Find a Number記憶化寬

A. Find a Number time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output

迷宮+火山

連結:https://ac.nowcoder.com/acm/contest/302/E 來源:牛客網   時間限制:C/C++ 1秒,其他語言2秒 空間限制:C/C++ 32768K,其他語言65536K 64bit IO Format: %lld 題目描述   &

HDU 2612 Find a way雙層BFS)

傳送門 題目大概的意思就是兩個人Y和M約在KFC見面,找到一家KFC使得兩個人到達時的距離之和最短,並且Y不能經過M的出發地,M不能經過Y的出發地。 思路也很清晰,用兩次BFS,找出Y和M分別到每個KFC所需要的時間,然後把兩個人到達每個KFC需要的時間加起來,找出最小值,這個最小值就是題目

Find a way 廣度優先

需要 sam namespace 比較 建立 繼續 tip total lag 題目: Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo o

HDU2612 Find a wayBFS簡單模板

Find a way Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 24558    Accepted Sub

圖 | 種遍歷方式:深度優先搜尋DFS、深和廣度優先搜尋BFS、

前邊介紹了有關圖的 4 種儲存方式,本節介紹如何對儲存的圖中的頂點進行遍歷。常用的遍歷方式有兩種:深度優先搜尋和廣度優先搜尋。 深度優先搜尋(簡稱“深搜”或DFS) 圖 1 無向圖 深度優先搜尋的過程類似於樹的先序遍歷,首先從例

HDU 2102 A計劃廣度優先搜尋+資料

A計劃 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 17222    Accepted Submissi

HDU 2612 Find a way 題解

Find a way Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 25379  &n

【hdu1241Oil Deposits】【HDU 2612 Find a way

HDU1241Oil Deposits 求聯通塊數量 Sample Input 1 1 * 3 5 *@*@* **@** *@*@* 1 8 @@****@* 5 5  ****@ *@@*@ *@**@ @@@*@ @@**@ 0 0  Sample Output 0 1

HDU 2612 Find a way

題目連結:傳送門 Problem Description Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifen

HDU-2612-Find a way

fin 捕獲 fir sizeof ret amp second using 時間 鏈接:https://vjudge.net/problem/HDU-2612#author=zhang95986 題意: hsj和lsh最近迷上了pokemon go的遊戲。在雙十一大物

HDU 6153 A Secret擴展kmp

出現的次數 include for %d target recommend lar out big A Secret Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 256000/256000 K (Java/

HDU 2102 A計劃簡單DFS

ont inpu clu 通過 csdn input tin 一次 != Description 可憐的公主在一次次被魔王擄走一次次被騎士們救回來之後,而今,不幸的她再一次面臨生命的考驗。魔王已經發出消息說將在T時刻吃掉公主,因為他聽信謠言說吃公主的肉也能長生不老。年邁的國

N - Find a way

mos inpu 標記 oid += expr them people 多次 Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one ye

ACM/ICPC 2018亞洲區預選賽北京賽站網絡賽 A.Saving Tang Monk II優先隊列

clas d+ ace char bool style monk per urn #include<bits/stdc++.h> using namespace std; const int maxN = 123; const int inf = 1e9 +

Deduplication on a Linked List個測試點

#include <stdio.h> #include <stdlib.h> typedef int ElementType; typedef struct LinkedNode *LinkedList; struct LinkedNode {

2015 ACM/ICPC Asia Regional Hefei Online I - Find a path推公式+dp

題目連結:傳送門   題意:在n*m的平面上,每個點都具有貢獻,問你從(1,1)到(n,m)找到一條路徑(只能往上或者往右),使得路徑上的點的方差最小。詢問最小的方差值為多少。   解決方法:首先根據簡化所給公式可以推出一個式子。推出式子的步驟如下 公式