#include<iostream>#include<cstdio>#include<queue>using namespace std;class P{public: char mark; int record; //搜索过这里没 int x,y; vector<int> XX; vector<int> Y...
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2021-07-12 10:17:14
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问题链接:HDU1026 Ignatius and the Princess I。题意简述:从矩阵的左上角走到右下角所需的最短时间,并且要求输出走的过程。矩阵中"."是可以走的,"X"是墙,n(数字1-9)是怪兽,需要战斗数字所示的时间。对于每个测试实例,先输入n和m...
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2016-08-17 03:09:00
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问题链接:HDU1026 Ignatius and the Princess I。题意简述:从矩阵的左上角走到右下角所需的最短时间,并且要求输出走的过程。矩阵中"."是可以走的,"X"是墙,n(数字1-9)是怪兽,需要战斗数字所示的时间。对于每个测试实例,先输入n和m...
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2016-08-17 03:09:00
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HDU1026 Ignatius and the Princess ITime Limit: 2000/1000 MS (Java/Others) Memory
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2022-11-30 13:12:41
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Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11710 Accepted Submi
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2017-05-23 17:16:00
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Problem Description The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166's castle. The castle is a large labyrinth. To make the problem simply, we assume the labyrinth is a N*M two-dimensional array which left-to
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2013-08-02 22:32:00
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Problem Description The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166's castle. The castle is a large labyrinth. To make the problem simply, we assume the labyrinth is a N*M two-dimensional array which left-to
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2013-08-03 22:40:00
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Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 18858 Accepted Submissi
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2022-08-09 22:21:20
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地址:http://acm.hdu.edu.cn/showproblem.php?pid=1026题意:n*m迷宫,求从(0,0)到(n-1,m-1)的最少时间。'X'是墙,'.'是空地,'1'-'9'表示有怪物,消灭之需要数字对应的时间。mark:最近一直在写搜索题。搜索题有的很费劲,但是写多也就觉得嗯,就那么回事。这题也没啥可说的,直接BFS。和最简单的BFS不同的是,判断一个过去到过的点是否可走,需要多开一个数组来记录,然后判断过去到达的时间和现在到达的时间的大小。(其实最好的方法是优先队列么。。。懒得写而已。)然后因为要求
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2012-01-17 05:28:00
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Ignatius and the Princess ITime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14212 Accepted Submission(s): 4488Special JudgeProblem
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2023-08-15 17:30:49
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原题链接: http://acm.hdu.edu.cn/showprob终点的最小秒,并输出路径。二:分析
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2022-12-06 19:19:02
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单纯bfs问题。
左上角是入口迷宫,右下角是出口,有战斗在迷宫1~9后卫,他们需要击败1~9时间。
遇到怪物就能判断是什么。步骤反向输出可。
因为需要记录的步骤,队列了。
也能够使用priority_queue 优化到 0ms,
模拟queue
#include<cstdio>
#include<cstring>
#include<string>
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2015-09-17 18:59:00
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上班之余抽点时间出来写写博文,希望对新接触的朋友有帮助。今天在这里和大家一起学习一下最小方向本来一道简略的搜索题,输出方式是难点每日一道理 在每个人心中,都曾停留过那些值得怀念的人,也许还在,也许早
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2013-05-28 23:28:00
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广搜的一个简单变形,思路还是一样的,依旧是维护一个队列,将一个节点不断的扩展,扩展完后出队。这道题还有两个特点就是:可能遇到怪兽,因此需要额外花费n秒的时间来打败它。最终还要输出路径。因此结构体里面prex 和 prey就是来记录下一个格子的坐标的。因为有了怪兽所以我们不能一搜到起点就退出搜索,因为...
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2014-08-13 23:16:00
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Ignatius and the Princess ITime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8405Accepted Submission(s): 2499 Special JudgeProblem DescriptionThe Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Prince
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2013-03-28 10:47:00
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题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1026在这博客学的 ttp://www.wutianqi.com/?p=2354感觉看了这个之后收获良多#include <queue>#include <stack>#include <cstdio>#include <iostream>#include <string.h>#include <algorithm>using namespace std;typedef struct node{ int x,y,dis; int
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2021-07-29 16:17:19
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Ignatius and the Princess I Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17171 Accepted Submis
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2021-07-21 15:47:34
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