题意: n*m的迷宫,有一些格能走("."),有一些格不能走("#").起始点为"@". 有K个物体.(K<=4),每个物体都是放在"."上. 问最少花多少步可以取完所有物体. 思路: BFS+状压,看代码. 代码: struct node{ int x,s; node(int _x,int _s){ x=_x, s=_s; } }; int n,m,k,sx,sy; char graph[105][105]; int…

Stealing Harry Potter's Precious Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 126    Accepted Submission(s): 63 Problem Description Harry Potter has some precious. For example, his invisible…

Stealing Harry Potter's Precious Problem Description Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in holiday, Harry Potter has to go back to uncle Vernon's home. But he can't bring his…

Stealing Harry Potter's Precious Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in…

2013杭州区域赛现场赛二水... 类似“胜利大逃亡”的搜索问题,有若干个宝藏分布在不同位置,问从起点遍历过所有k个宝藏的最短时间. 思路就是,从起点出发,搜索到最近的一个宝藏,然后以这个位置为起点,搜索下一个最近的宝藏,直至找到全部k个宝藏.有点贪心的感觉. 由于求最短时间,BFS更快捷,但耗内存,这道题就卡在这里了... 这里记录了我几次剪枝的历史...题目要求内存上限32768KB,就差最后600KB了...但我从理论上觉得已经不能再剪了,留下的结点都是盲目式搜索必然要访问的结点. 在此贴…

状压BFS 注意在用二维字符数组时,要把空格.换行处理好. #include<stdio.h> #include<algorithm> #include<string.h> #include<queue> using namespace std; #define INF 0x3f3f3f3f int sx,sy,C,n,m; int ans; ][][<<]; ][]; ][]; ,,-,},dy[]={,,,-}; <=a&&am…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4771 题目意思:'@'  表示的是起点,'#' 表示的是障碍物不能通过,'.'  表示的是路能通过的: 目的:让你从 '@' 点出发,然后每个点只能走一次,求出最小的距离: 解题思路:先用 bfs 求解出任意两点之间的距离,用 ans[i][j],表示点 i 到点  j 的距离: 然后用 dfs 递归求出从起点经过所有点的距离中,比较出最小的: AC代码: #include<iostream> #…

题目链接:hdu 4771 Stealing Harry Potter's Precious 题目大意:在一个N*M的银行里,贼的位置在'@',如今给出n个宝物的位置.如今贼要将全部的宝物拿到手.问最短的路径,不须要考虑离开. 解题思路:由于宝物最多才4个,加上贼的位置,枚举两两位置,用bfs求两点距离,假设存在两点间不能到达,那么肯定是不能取全然部的宝物. 然后枚举取宝物的顺序.维护ans最小. #include <cstdio> #include <cstring> #incl…

Problem Description Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in holiday, Harry Potter has to go back to uncle Vernon's home. But he can't bring his precious with him. As you know,…

http://acm.hdu.edu.cn/showproblem.php?pid=5025 Saving Tang Monk Problem Description   <Journey to the West>(also <Monkey>) is one of the Four Great Classical Novels of Chinese literature. It was written by Wu Cheng'en during the Ming Dynasty.…

题目:给出一个二维图,以及一个起点,m个中间点,求出从起点出发,到达每一个中间的最小步数. 思路:由于图的大小最大是100*100,所以要使用bfs求出当中每两个点之间的最小距离.然后依据这些步数,建立一个新的图,使用dfs求出最佳步数. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <queue> #define INF 100000000 using names…

Stealing Harry Potter's Precious Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1343    Accepted Submission(s): 642 Problem Description Harry Potter has some precious. For example, his invisib…

题意: n*m的迷宫.多多要从(1,1)到达(n,m).每移动一步消耗1秒.有P种钥匙. 有K个门或墙.给出K个信息:x1,y1,x2,y2,gi    含义是(x1,y1)与(x2,y2)之间有gi.gi=0:墙   1,2,3.... :第1种门,第2种门,第3种门..... 有S把钥匙.给出S个信息:x1,y1,qi    含义是位置(x1,y1)上有一把第qi种的钥匙. 问多多最少花多少秒到达(n,m).若无法到达输出-1. 数据范围: (1<= n, m <=50, 0<= p…

方格取数(2) Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6206    Accepted Submission(s): 1975 Problem Description 给你一个m*n的格子的棋盘,每个格子里面有一个非负数. 从中取出若干个数,使得任意的两个数所在的格子没有公共边,就是说所取数所在的2个格子不能相邻,并且取出的…

https://loj.ac/problem/6121 BFS + 状压 写过就好想,注意细节debug #include <bits/stdc++.h> #define read read() #define up(i,l,r) for(register int i = (l);i <= (r);i++) #define down(i,l,r) for(register int i = (l);i >= (r);i--) #define traversal_vedge(i) fo…

Description Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in holiday, Harry Potter has to go back to uncle Vernon's home. But he can't bring his precious with him. As you know, uncle Ve…

http://acm.hdu.edu.cn/showproblem.php?pid=2209 不知为啥有种直觉.会出状压+搜索的题,刷几道先 简单的BFS.状压表示牌的状态, //#pragma comment(linker, "/STACK:102400000,102400000") #include <cstdio> #include <cstring> #include <algorithm> #include <string> #…

http://acm.hdu.edu.cn/showproblem.php?pid=5025 N*N矩阵 M个钥匙 K起点,T终点,S点需多花费1点且只需要一次,1-9表示9把钥匙,只有当前有I号钥匙才能拿I+1号钥匙,可以不拿钥匙只从上面走过 BFS+优先队列.蛇最多只有5条,状压即可. #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <…

题目链接 The input contains mutiple testcases. Please process till EOF.For each testcase, the first line contains two integers N (1 ≤ N ≤ 15), the side length of the square map and M (1 ≤ M ≤ 15), the number of tunnels.The map of the city is given in the…

题意:这次魔王汲取了上次的教训,把Ignatius关在一个n*m的地牢里,并在地牢的某些地方安装了带锁的门,钥匙藏在地牢另外的某些地方.刚开始 Ignatius被关在(sx,sy)的位置,离开地牢的门在(ex,ey)的位置.Ignatius每分钟只能从一个坐标走到相邻四个坐标中的其中一 个.魔王每t分钟回地牢视察一次,若发现Ignatius不在原位置便把他拎回去.经过若干次的尝试,Ignatius已画出整个地牢的地图.现在请你帮 他计算能否再次成功逃亡.只要在魔王下次视察之前走到出口就算离开地牢…

题意: 给出n个资源串,m个病毒串,现在要如何连接资源串使得不含病毒串(可以重叠,样例就是重叠的). 题解: 这题的套路和之前的很不同了,之前的AC自动机+DP的题目一般都是通过teir图去转移, 这题有点小不同,认真分析这一题,我们可以知道n个资源串,m个病毒串在AC自动机的上面是哪一个节点 所以可以根据teir图去处理出每一个资源串的节点不经过病毒串节点的最短距离,这个可以通过BFS实现. 这个一开始要处理AC自动机上的root节点,毕竟是从root节点开始走的. 处理出了每个点到其他点的最…

这是今天下午的互测题,只得了60多分 分析一下错因: $dis[i][j]$只记录了相邻的两个岛屿之间的距离,我一开始以为可以,后来$charge$提醒我有可能会出现来回走的情况,而状压转移就一次,无法实现来回走的情况,所以加了一个类似$floyed算法$的三重循环来更新每个点的距离,然后状态转移就可以了,枚举起点和终点,最后统计答案 #include<cstdio> #include<cstring> #include<algorithm> using namespa…

Problem Description Bob is travelling in Xi’an. He finds many secret tunnels beneath the city. In his eyes, the city is a grid. He can’t enter a grid with a barrier. In one minute, he can move into an adjacent grid with no barrier. Bob is full of cur…

孤岛营救问题 https://www.luogu.org/problemnew/show/P4011 用状压DP标记拿到钥匙的数量 #include<iostream> #include<cstring> #include<string> #include<algorithm> #include<cmath> #include<queue> #include<vector> #include<cstdio> u…

Arseny likes to organize parties and invite people to it. However, not only friends come to his parties, but friends of his friends, friends of friends of his friends and so on. That's why some of Arseny's guests can be unknown to him. He decided to…

来自xjy的签到题   Description 爱丽丝冒险来到了红皇后一个n*n大小的花园,每个格子由'.'或'#'表示,'.'表示爱丽丝可以到达这个格子,‘#’表示爱丽丝不能到达这个格子,爱丽丝每1分钟可以移动到非'#'的相邻格子(与当前所在格子具有公共边).花园下面有m个隧道,每个隧道有一个出口和一个入口.当爱丽丝到达隧道的入口时,她可以选择(也可以不选择)进入隧道入口,并通过隧道一次,然后立即(不花费时间)出现在隧道出口.爱丽丝一开始可以降临在花园的任何地方.有好奇心的爱丽丝想知道,她通过…

Problem Description Rompire is a robot kingdom and a lot of robots live there peacefully. But one day, the king of Rompire was captured by human beings. His thinking circuit was changed by human and thus became a tyrant. All those who are against him…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5977 题解:这题一看就知道是状压dp然后看了一下很像是点分治(有点明显)然后就是简单的点分治+状压dp,这里只要稍微改一下模版就行了.还有注意一下这里的cau状态枚举然后就没什么了 #include <iostream> #include <cstring> #include <cstdio> using namespace std; typedef long long l…

题目大意: 输入w h,接下来输入h行w列的图 ' . ':干净的点:  ' * ' :垃圾:  ' x ' : 墙:  ' o ' : 初始位置: 输出 清理掉所有垃圾的最短路径长度 无则输出-1 Sample Input 7 5........o...*.........*...*........15 13.......x..........o...x....*.........x..............x..............x......................xxxx…

练习dfs和bfs的好题. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<algorithm> #include<…