Collect More Jewels Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6713    Accepted Submission(s): 1558 Problem Description It is written in the Book of The Lady: After the Creation, the cruel…

http://acm.hdu.edu.cn/showproblem.php?pid=1044 代码题,没什么好说的,先预处理出两点间距离,然后dfs搜一下找最大值 #include <iostream> #include <cstdio> #include <algorithm> #include <queue> using namespace std; const int INF=0xfffffff; int n,m,L,M,ans,sum; ][]; ]…

Collect More Jewels Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5345    Accepted Submission(s): 1189 Problem Description It is written in the Book of The Lady: After the Creation, the cruel…

非常经典的一类题型 没有多个出口.这里题目没有说清楚 Collect More Jewels Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4684    Accepted Submission(s): 983 Problem Description It is written in the Book of The Lady: Afte…

Collect More Jewels Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6148    Accepted Submission(s): 1386 Problem Description It is written in the Book of The Lady: After the Creation, the cruel…

Collect More Jewels Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7543    Accepted Submission(s): 1761 Problem Description It is written in the Book of The Lady: After the Creation, the cruel…

Collect More Jewels Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6739    Accepted Submission(s): 1564 Problem Description It is written in the Book of The Lady: After the Creation, the cruel…

题意: 一个n*m的迷宫,在t时刻后就会坍塌,问:在逃出来的前提下,能带出来多少价值的宝藏. 其中: ’*‘:代表墙壁: '.':代表道路: '@':代表起始位置: '<':代表出口: 'A'~'J':代表宝藏: 解题思路: 本题有两种思路:1,bfs+状态压缩:相比之下耗时多: 2,bfs+dfs:耗时较少: 在这里着重介绍第一种方法:(博主也是现学现卖啦^_^) 因为宝藏的数目比较少,所以我们可以对宝藏进行状态压缩(用二进制表示宝藏是否已经拾取,1表示已拾取,0表示未拾取), 标记数组vis…

题意:              给你起点,终点,图上有墙有路还有宝物,问你在规定时间内能否能到终点,如果能问最多能捡到多少宝物. 思路:           看完这个题目果断 BFS+三维的mark[][][] 第三维用二进制压缩的思想去解决,结果TLE了,我后来在网上看了看,发现有人用二进制压缩ac了,这更坚定了我的决心啊,不停的优化 超时 优化 超时 就这样我超时了 50多次,我tm恶心了,直接粘了个网上说和我想法一样的代码交上去了,结果 TLE 了, 我 fuck ,超时了早说啊... …

BFS + 状态压缩 险过 这个并不是最好的算法 但是写起来比较简单 , 可以AC,但是耗时比较多 下面是代码 就不多说了 #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <queue> using namespace std; #define Max(a,b) (a>b?a:b) #define Min(a,b) (a…