POJ:Dungeon Master(BFS模板题)】的更多相关文章

Description You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot m…
You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonal…
Sliding Window POJ - 2823 单调队列模板题 题意 给出一个数列 并且给出一个数m 问每个连续的m中的最小\最大值是多少,并输出 思路 使用单调队列来写,拿最小值来举例 要求区间最小值 就是维护一个单调递增的序列 对于样例 8 3 1 3 -1 -3 5 3 6 7 我们先模拟一遍 1.队列为空 1 进队 队列:1 2.3>队尾元素 3 进队 队列: 1 3 3.-1小于队尾元素,一直从尾部出队知道找到比-1小的元素或者队列为空 队列:-1 当队列中元素大于m的时候从队头删…
Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16748   Accepted: 6522 Description You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled…
题目链接:http://poj.org/problem?id=2251 Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37296   Accepted: 14266 Description You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit…
Dungeon Master Description You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or dow…
<题目链接> 题目大意: 给定一颗树,求出树的直径. 解题分析:树的直径模板题,以下程序分别用树形DP和两次BFS来求解. 树形DP: #include <cstdio> #include <algorithm> using namespace std; ; struct Edge{ int to,val,nxt; Edge(,,):to(_to),val(_val),nxt(_nxt){} }e[N<<]; int n,m,cnt,ans; int dp1…
Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17555   Accepted: 6835 Description You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled…
题目链接:http://icpc.njust.edu.cn/Problem/Hdu/1242/ 这次的迷宫是有守卫的,杀死一个守卫需要花费1个单位的时间,所以以走的步数为深度,在每一层进行搜索,由于走一步的花费不一定是1,所以我们需要用优先队列寻找最优值.这个题目真是模板题. 代码如下: #include<bits/stdc++.h> using namespace std; typedef unsigned int ui; typedef long long ll; typedef unsi…
一开始预习是百度的算法 然后学习了一下 然后找到了学长的ppt 又学习了一下.. 发现..居然不一样... 找了模板题试了试..百度的不好用 反正就是wa了..果然还是应当跟着学长混.. 图两边的点分别是行数和列数 每有一个点 就让所处行列连一条边 求最小点覆盖 然后卡住...后来看了增林的博客... 最小点覆盖=最大匹配数 果然是模板题.. 然后wa.. 后来发现是当进行对左边点的遍历的时候 每次都要mem一次vis数组 应该是每次找之前都重新清空啊..不然下次怎么找啊...增光路对点的是否被…