Description Captain Clearbeard decided to go to the harbour for a few days so his crew could inspect and repair the ship. Now, a few days later, the pirates are getting landsick(Pirates get landsick when they don't get enough of the ships' rocking mo…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4198 Quick out of the Harbour Description Captain Clearbeard decided to go to the harbour for a few days so his crew could inspect and repair the ship. Now, a few days later, the pirates are getting land…

Quick out of the Harbour Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1441    Accepted Submission(s): 575 Problem DescriptionCaptain Clearbeard decided to go to the harbour for a few days so…

题目链接:hdu4198 题目大意:求起点S到出口的最短花费,其中#为障碍物,无法通过,‘.’的花费为1 ,@的花费为d+1. 需注意起点S可能就是出口,因为没考虑到这个,导致WA很多次....... #include<stdio.h> #include<string.h> #include<algorithm> #include<queue> using namespace std; char map[505][505]; int d[4][2] = {{…

题目链接:pid=2102">http://acm.hdu.edu.cn/showproblem.php?pid=2102 这道题属于BFS+优先队列 開始看到四分之中的一个的AC率感觉有点吓人,后来一做感觉就是模板改了点东西而已,一遍就AC了,只是在主函数和全局变量里面都定义了n和m导致我白白浪费了debug的时间. 果然全局变量得小心用啊. 跟模板一样的,定义一个结构体,仅仅只是多加了个參数,就是迷宫的层数,我用0代表第一层.1代表第二层,这在数组里面会体现的. struct node…

找到朋友的最短时间 Sample Input7 8#.#####. //#不能走 a起点 x守卫 r朋友#.a#..r. //r可能不止一个#..#x.....#..#.##...##...#.............. Sample Output13 bfs+优先队列 #include <iostream> #include <cstring> #include <cstdio> #include <queue> using namespace std;…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1548 There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if yo…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1026 Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21841    Accepted Submission(s): 7023Special Judge Problem Descrip…

HDU 1312 题目大意: 一个地图里面有三种元素,分别为"@",".","#",其中@为人的起始位置,"#"可以想象为墙,然后.为可以走的空地,求人可以走的最大点数. 解题思路:从起点开始,起点的四个方向满足条件的点分别入队(放置重复入队,需只要一入队就标记已访问而不是取出时再进行标记),直至队为空. /*HDU 1312 ----- Red and Black 入门搜索 BFS解法*/ #include <cstd…

Resource Archiver Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 100000/100000 K (Java/Others)Total Submission(s): 2382    Accepted Submission(s): 750 Problem Description Great! Your new software is almost finished! The only thing left to…