题意:给一个地图,孙悟空(K)救唐僧(T),地图中'S'表示蛇,第一次到这要杀死蛇(蛇最多5条),多花费一分钟,'1'~'m'表示m个钥匙(m<=9),孙悟空要依次拿到这m个钥匙,然后才能去救唐僧,集齐m个钥匙之前可以经过唐僧,集齐x个钥匙以前可以经过x+1,x+2..个钥匙,问最少多少步救到唐僧. 解法:BFS,每个节点维护四个值: x,y : 当前坐标 key :已经集齐了key个钥匙 step:已经走了多少步 S :   蛇的访问状态 (2^5的数表示,某位为1表示已经杀过了) 然后把唐僧…

作者:jostree 转载请注明出处 http://www.cnblogs.com/jostree/p/4092939.html 题目链接:hdu 5025 Saving Tang Monk 状态压缩dp+广搜 使用dp[x][y][key][s]来记录孙悟空的坐标(x,y).当前获取到的钥匙key和打死的蛇s.由于钥匙具有先后顺序,因此在钥匙维度中只需开辟大小为10的长度来保存当前获取的最大编号的钥匙即可.蛇没有先后顺序,其中s的二进制的第i位等于1表示打死了该蛇,否则表示没打死.然后进行广度…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=5025 Saving Tang Monk Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 3242    Accepted Submission(s): 1127 Problem Description <Journey to the West>…

Saving Tang Monk Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 941    Accepted Submission(s): 352 Problem Description <Journey to the West>(also <Monkey>) is one of the Four Great Clas…

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. In this novel, Monkey King Sun Wukong, pig Zhu Bajie and Sha Wujing, escor…

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. In this novel, Monkey King Sun Wukong, pig Zhu Bajie and Sha Wujin…

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. In this novel, Monkey King Sun Wukong, pig Zhu Bajie and Sha Wujin…

/* 这是我做过的一道新类型的搜索题!从来没想过用四维数组记录状态! 以前做过的都是用二维的!自己的四维还是太狭隘了..... 题意:悟空救师傅 ! 在救师父之前要先把所有的钥匙找到! 每种钥匙有 k 种, 每一种有多个! 只要求找到每一种的其中一个就可以! 找钥匙的顺序按照 第1种, 第2种, 第3种 ....第k种! 找钥匙的时间是一步, 走到相邻空地的时间是一步, 打蛇的时间就是两步! 求找到师傅的最少步数! 这里说一下 state[N][N][10][35]表示的含义: ->state[…

#include<bits/stdc++.h> using namespace std; ; ; char G[maxN][maxN], snake[maxN][maxN]; ]; int n, m, sx, sy, ex, ey, ans; ][] = {{,},{,},{,-},{-,}}; struct node { int x, y, t, key, killed; //坐标, 步数, 已经拿到的钥匙, 有没有杀蛇 bool operator < (const node&…

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.…

Saving Tang Monk II https://hihocoder.com/problemset/problem/1828 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 <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…

题目1 : Saving Tang Monk II 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 <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. In this novel, Monkey…

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…

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…

做法:优先队列模板题,按步数从小到大为优先级,PASS掉曾经以相同氧气瓶走过的地方就好了 题目1 : Saving Tang Monk II 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 <Journey to the West>(also <Monkey>) is one of the Four Great Classical Novels of Chinese literature. It was written by Wu Cheng'en du…

任意门:http://hihocoder.com/problemset/problem/1828 Saving Tang Monk II 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 <Journey to the West>(also <Monkey>) is one of the Four Great Classical Novels of Chinese literature. It was written by Wu Cheng'en duri…

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…

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 <…

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> #…

描述 <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. In this novel, Monkey King Sun Wukong, pig Zhu Bajie and Sha Wujing, escorted Tang…

题目链接 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. In this novel, Monkey King Sun Wukong, pig Zhu Bajie and Sha…

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…

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…

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…

Paint on a Wall Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 830    Accepted Submission(s): 325 Problem Description Annie wants to paint her wall to an expected pattern. The wall can be repr…

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…

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…

题目链接:https://cn.vjudge.net/problem/HDU-1565 Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description 给你一个n*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,就是说所取的数所在的2个格子不能相邻,并且取出的数的和最大.   Input 包括多个测试实例,每…

<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. In this novel, Monkey King Sun Wukong, pig Zhu Bajie and Sha Wujing, escorted Tang Mon…

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…