http://acm.hdu.edu.cn/showproblem.php?pid=1242 问题:牢房里有墙(#),警卫(x)和道路( . ),天使被关在牢房里位置为a,你的位置在r处,杀死一个警卫要一秒钟,每走一步要一秒钟,求最短时间救出天使,不能救出则输出:Poor ANGEL has to stay in the prison all his life.  求最短路径,果断广搜BFS 限制及剪枝: 1.墙不能走,不能离开牢房范围 2.杀死一个警卫要多花一秒钟 3.当前步骤大于等于最短时间…

题目链接:hdu 1242 这题也是迷宫类搜索,题意说的是 'a' 表示被拯救的人,'r' 表示搜救者(注意可能有多个),'.' 表示道路(耗费一单位时间通过),'#' 表示墙壁,'x' 代表警卫(耗费两个单位时间通过),然后求出 'r' 能找到 'a' 的最短时间,找不到输出 "…………"(竟然在这里也 wa 了一发 -.-||).很明显是广搜了,因为 'r' 可能有多个,所以我们反过来从 'a' 开始搜,每次搜到 'r' 都更新最小时间值(很重要的一个转换!).可是这题因为通过 '…

Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 14144    Accepted Submission(s): 6058 Problem Description At the entrance to the university, there is a huge rectangular billboard of…

http://acm.hdu.edu.cn/showproblem.php?pid=1242 感觉题目没有表述清楚,angel的朋友应该不一定只有一个,那么正解就是a去搜索r,再用普通的bfs就能过了. 但是别人说要用优先队列来保证时间最优,我倒是没明白,步数最优跟时间最优不是等价的吗?就算士兵要花费额外时间,可是既然先到了目标点那时间不也一定是最小的? 当然用优先队列+ a去搜索r是最稳妥的. #include <cstdio> #include <cstring> #inclu…

此刻再看优先队列,不像刚接触时的那般迷茫!这也许就是集训的成果吧! 加油!!!优先队列必须要搞定的! 这道题意很简单!自己定义优先级别! +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ =================================================================================== +++++++++++++…

题目来源: http://acm.hdu.edu.cn/showproblem.php?pid=1242 题目描述: Problem Description   Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the pris…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1242 题目描述: Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison…

题目链接:Rescue 进度落下的太多了,哎╮(╯▽╰)╭,渣渣我总是埋怨进度比别人慢...为什么不试着改变一下捏.... 開始以为是水题,想敲一下练手的,后来发现并非一个简单的搜索题,BFS做肯定出事...后来发现题目里面也有坑 题意是从r到a的最短距离,"."相当时间单位1,"x"相当时间单位2,求最短时间 HDU 搜索课件上说,这题和HDU1010相似,刚開始并没有认为像剪枝,就改用  双向BFS   0ms  一Y,爽! 网上查了一下,神牛们居然用BFS+优…

Rescue Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 14   Accepted Submission(s) : 7 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Angel was caught by the MOLIGPY…

Rescue Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12441 Accepted Submission(s): 4551 Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is descri…

http://acm.hdu.edu.cn/showproblem.php?pid=4608 听说这个题是比赛的签到题......无语..... 问题:给你一个数x,求比它大的数y. y的要求: 1.y>x 2.y的每一位数相加的和为10的倍数 3.求最小的y 直接模拟,个位数加一然后求各位数总和是否为10的倍数... 有的人还考虑了前导零和后导零导致错误.这个题不用考虑那么多...坑.... 这个题一开始做的好郁闷,没有考虑到 最高位进位,导致我提交全是WA,所以我用的数组存数据,而且是倒着存…

HDU 1241 是深搜算法的入门题目,递归实现. 原题目传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1241 代码仅供参考,c++实现: #include <iostream> using namespace std; ][]; int p,q; void dfs(int x,int y){ land[x][y] = '*'; ][y]!= ][y] != ] != ] != ][y+]!= ][y-] != ][y-] != ][y+] !…

题目链接 ZOJ链接 Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison. Angel's friends want to save Angel. Their task…

Rescue Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12927    Accepted Submission(s): 4733 Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is d…

Rescue Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24205    Accepted Submission(s): 8537 Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is d…

数塔 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 36261    Accepted Submission(s): 21659 Problem Description 在讲述DP算法的时候,一个经典的例子就是数塔问题,它是这样描述的: 有如下所示的数塔,要求从顶层走到底层,若每一步只能走到相邻的结点,则经过的结点的数字之和最大是多少?…

题目 /******************以下思路来自百度菜鸟的程序人生*********************/ bfs即可,可能有多个’r’,而’a’只有一个,从’a’开始搜,找到的第一个’r’即为所求 需要注意的是这题宽搜时存在障碍物,遇到’x’点是,时间+2,如果用普通的队列就 并不能保证每次出队的是时间最小的元素,所以要用优先队列,第一次用优先队列,还不熟练哇 优先队列(priority_queue)的基本操作: empty(); 队列为空返回1 pop();   出队 push(…

Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison. Angel's friends want to save Angel. Their task is: approa…

Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison. Angel's friends want to save Angel. Their task is: approa…

欢迎"热爱编程"的高考少年--报考杭州电子科技大学计算机学院 Just a Hook Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 20889    Accepted Submission(s): 10445 Problem Description In the game of DotA, Pudge's meat hook…

Milk Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15697    Accepted Submission(s): 3947 Problem Description Ignatius drinks milk everyday, now he is in the supermarket and he wants to choose…

题意: 一个天使a被关在迷宫里,她的很多小伙伴r打算去救她.求小伙伴就到她须要的最小时间.在迷宫里有守卫.打败守卫须要一个单位时间.假设碰到守卫必须要杀死他 思路: 天使仅仅有一个,她的小伙伴有非常多,所以能够让天使找她的小伙伴,一旦找到小伙伴就renturn.时间小的优先级高.优先队列搞定 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include&l…

题意:X代表卫兵,a代表终点,r代表起始点,.代表路,#代表墙,走过.要花费一秒,走过x要花费2秒,求从起点到终点的最少时间. 析:一看到样例就知道是BFS了吧,很明显是最短路径问题,不过又加了一个条件——时间,所以我们用优先队列去优先获取时间短的路径,总体实现起来没有太大难度. 代码如下: #include <iostream> #include <cstdio> #include <vector> #include <set> #include <…

题目链接 Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison. Angel's friends want to save Angel. Their task is: a…

第一次用容器做的BFS题目,题目有个地方比较坑,就是遍历时的方向,比如上下左右能AC,右上左下就WA #include <stdio.h> #include <string.h> #include <iostream> #include <queue> using namespace std; char map[205][205]; int x_begin,y_begin,flag,n,m; int v[205][205],d[4][2] = { {-1,0…

Design T-Shirt Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6527    Accepted Submission(s): 3061 Problem Description Soon after he decided to design a T-shirt for our Algorithm Board on Free…

杭州人称那些傻乎乎粘嗒嗒的人为62(音:laoer). 杭州交通管理局经常会扩充一些的士车牌照,新近出来一个好消息,以后上牌照,不再含有不吉利的数字了,这样一来,就可以消除个别的士司机和乘客的心理障碍,更安全地服务大众. 不吉利的数字为所有含有4或62的号码.例如: 62315 73418 88914 都属于不吉利号码.但是,61152虽然含有6和2,但不是62连号,所以不属于不吉利数字之列. 你的任务是,对于每次给出的一个牌照区间号,推断出交管局今次又要实际上给多少辆新的士车上牌照了.  In…

Catch That Cow Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8999    Accepted Submission(s): 2837 Problem Description Farmer John has been informed of the location of a fugitive cow and wants…

验证角谷猜想 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6653    Accepted Submission(s): 3417 Problem Description 数论中有很多猜想尚未解决,当中有一个被称为"角谷猜想"的问题,该问题在五.六十年代的美国多个著名高校中曾风行一时,这个问题是这样描写叙述的:不论什么一个…

Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison. Angel's friends want to save Angel. Their task is: approach Angel…