题意: 求把S和所有的A连贯起来所用的线的最短长度... 这道题..不看discuss我能wa一辈子... 输入有坑... 然后,,,也没什么了...还有注意 一次bfs是可以求当前点到所有点最短距离的... #include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <algorithm> #include <cmath> #d…

Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9821   Accepted: 3283 Description The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg collective is the term used to describe the gr…

Borg Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12032   Accepted: 3932 Description The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg collective is the term used to desc…

题意:从S出发,去抓每一个A,求总路径最短长度.在S点和A点人可以分身成2人,不过一次只能让一个人走. 思路是先利用BFS求出各点之间的距离,建成图,再套用最小生成树模板. 一次性A了.不过觉得在判断第几个编号的点时稍显麻烦了. #include <iostream> #include <cstdio> #include <cstring> #include <queue> using namespace std; const int inf=1000000…

链接: http://poj.org/problem?id=3026 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=82831#problem/J Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10713   Accepted: 3559 Description The Borg is an immensely powerful race of enhance…

在一个迷宫里面需要把一些字母.也就是 ‘A’ 和 ‘B’连接起来,求出来最短的连接方式需要多长,也就是最小生成树,地图需要预处理一下,用BFS先求出来两点间的最短距离, ********************************************************************************** #include<algorithm> #include<stdio.h> #include<; ][] = { {,},{,},{-,},{,…

Description The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg collective is the term used to describe the group consciousness of the Borg civilization. Each Borg individual is linked to the c…

Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9718   Accepted: 3263 Description The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg collective is the term used to describe the gr…

题目链接:http://poj.org/problem?id=3026 题目大意:在一个y行 x列的迷宫中,有可行走的通路空格’  ‘,不可行走的墙’#’,还有两种英文字母A和S,现在从S出发,要求用最短的路径L连接所有字母,输出这条路径L的总长度. 解题思路:相当于所有的字母A和S都是结点,求连接这些结点的最小距离和,是最小生成树的题目.先用BFS求各点间的距离,然后再用Prim(Kruskal也可以)求出最小距离就可以了. 注意:输完行列m和n之后,后面有一堆空格,要用gets()去掉,题目…

(￣▽￣)" #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include<queue> using namespace std; const int INF=10e8; ; int col,row,k,minn; char str[MAXN][MAXN]; ][MAXN*],lowdis…