http://poj.org/problem?id=3026 #include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #define maxn 100 #define maxn1 100000 #include<queue> using namespace std; char s[maxn][maxn]; <<; int n,m; ][maxn…

Maze Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 3183   Accepted: 996 Description Acm, a treasure-explorer, is exploring again. This time he is in a special maze, in which there are some doors (at most 5 doors, represented by 'A', 'B…

题意:给定迷宫,可以更改高度比,问如何使最短路等于输入数据. 思路:由于是单调的,可以用二分答案,然后BFS验证.这里用优先队列,每次压入也要进行检查(dis大小)防止数据过多,A*也可以.好久不写图论,WA成狗 #include <iostream> #include <string> #include <cstring> #include <cstdio> #include <algorithm> #include <memory>…

题意,给你一个l,和一个地图,让你从起点走到终点,使得路程刚好等于l. 你可以选择一个系数,把纵向的地图拉伸或收缩,比如你选择系数0.5,也就是说现在上下走一步消耗0.5的距离,如果选择系数3,也就是说上下一步消耗3的距离. 左右不能改变. Hint中提示了答案在0--10之间,其实就透露出了二分的思想. 我们对系数P进行二分,对每一个系数P进行一次bfs,如果可以在小于等于l的步数内找到解,则增加下界,否则减小上界. 由于上下和左右的消耗值不相同,所以我们采用A*算法,设估价值为当前点到目标点…

#include <iostream> #include <stdio.h> #include <cstring> #define Max 0x7f7f7f7f using namespace std; ][]; ][]; ][]={{,},{-,},{,-},{,}}; ][]; ]; struct node { int x; int y; }; node path[]; void output(int head) { int tmp=pre[head]; ) { p…

走迷宫拿宝藏,拿到所有对应的钥匙才能开门 *解法:从起点bfs,遇到门时先放入队列中,取出的时候看钥匙够不够决定开不开门,如果不够就把它再放回队列继续往下走,当队列里只有几个门循环的时候就可以退出,所以记一个T<400 #include <iostream> #include <cstdio> #include <cstdlib> #include <queue> using namespace std; #define INF 1e9+10 char…

BFS广搜题目有时间一个个做下来 2009-12-29 15:09 1574人阅读 评论(1) 收藏 举报 图形graphc优化存储游戏 有时间要去做做这些题目,所以从他人空间copy过来了,谢谢那位大虾啦. pku 1175 Starry Night 题目地址:http://acm.pku.edu.cn/JudgeOnline/problem?id=1175 解法:BFS,要注意的是如何判断图形是一样的,我的做法就是计算每两个点的距离之和. 看:http://hi.baidu.com/doxi…

Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem 10983 18765 Y 1036 [ZJOI2008]树的统计Count 5293 13132 Y 1588 [HNOI2002]营业额统计 5056 13607 1001 [BeiJing2006]狼抓兔子 4526 18386 Y 2002 [Hnoi2010]Bounce 弹飞绵羊 43…

http://poj.org/problem?id=3026 Borg Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12086   Accepted: 3953 Description The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. The Borg coll…

Borg Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6634   Accepted: 2240 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 descr…

http://poj.org/problem?id=3026 Borg Maze Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3026 Description The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of…

题目链接:http://poj.org/problem?id=3026. 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. Ea…

题目链接:http://poj.org/problem?id=3026 感觉英语比题目本身难,其实就是个最小生成树,不过要先bfs算出任意两点的权值. #include <stdio.h> #include <string.h> #include <queue> #include <algorithm> using namespace std; ][]; ]; struct Point { int x, y; }point[]; struct Edge {…

题目链接:http://poj.org/problem?id=3026 Svenskt Masterskap我程序员/ Norgesmesterskapet 2001 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 g…

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…

链接: http://poj.org/problem?id=3026 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/J Borg Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6905   Accepted: 2315 Description The Borg is an immensely powerful race o…

id=3026">[POJ 3026]Borg Maze 一个考察队搜索alien 这个考察队能够无限切割 问搜索到全部alien所须要的总步数 即求一个无向图 包括全部的点而且总权值最小(最小生成树 BFS+最小生成树 Prim/Kruskal-懒死了 就这么贴吧--凑活看(￣┰￣*) #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #i…

Borg Maze 题意: 题目我一开始一直读不懂.有一个会分身的人,要在一个地图中踩到所有的A,这个人可以在出发地或者A点任意分身,问最少要走几步,这个人可以踩遍地图中所有的A点. 思路: 感觉就算读懂了题目,也比较难想到这用到了最小生成树的知识,因为可以分身,所以每个点可以向其他点都连上边.可以用bfs预处理出所有的S点,A点的连线,再跑一遍最小生成树,即可得出答案.这里有几点注意,一开始我bfs没有记录step,而是直接找到一点后算曼哈顿距离,这是不对的,因为可能是绕了一个圈到了这个点.还…

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…

Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9971   Accepted: 3347 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…

点击打开链接 Borg Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7097   Accepted: 2389 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 t…

Borg Maze 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/I 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 consciou…

Borg Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7998   Accepted: 2675 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 descr…

有几个错误,调试了几个小时,样例过后 1Y. 题目:http://poj.org/problem?id=3026 题意:就是让求A们和S的最小生成树 先用bfs找每两点的距离,再建树.没剪枝 63MS. #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #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…

链接:poj 3026 题意:y行x列的迷宫中,#代表阻隔墙(不可走).空格代表空位(可走).S代表搜索起点(可走),A代表目的地(可走),如今要从S出发,每次可上下左右移动一格到可走的地方.求到达全部的A的路线总距离最小值 分析:能够先用bfs从上下左右四个方向将全部的A,S两两之间的最短距离,题目的目的是将S与全部的A连通,使得总距离最小,所以任选一点開始按最小生成树的算法做即可,并不是非要从S点開始 注:题目输入x,y后可能有非常多空格,能够用gets将多余的空格取走,开数组是尽量开大点.…

(￣▽￣)" #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…

https://vjudge.net/problem/POJ-3026 题意 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. 分析 注意审题!!一开始读成可在任意位置分头走,结果题意限制了只能在字母处分开.这样就好做多了,L就是最小生成树的权值.因为这和生成树的处理过程很像,每一次找最近的点加入到集合中,即走过的路不用再考虑.本题难度在建图,先找出那几个字母的位置,给他们标号,…

http://poj.org/problem?id=3026 题意:给你一个迷宫,里面有 ‘S’起点,‘A’标记,‘#’墙壁,‘ ’空地.求从S出发,经过所有A所需要的最短路.你有一个特殊能力,当走到S或A时可以分身出任意多个人一起走.计算路程时就是所有人的总路程之和. 题解:想一下,是裸的最短路套上bfs. 先暴力bfs出各个点之间的距离,存边 然后kruskal #define _CRT_SECURE_NO_WARNINGS #include<stdio.h> #include<qu…

poj 2096 题目:http://poj.org/problem?id=2096 f[ i ][ j ] 表示收集了 i 个 n 的那个. j 个 s 的那个的期望步数. #include<cstdio> #include<cstring> #include<algorithm> #define db double using namespace std; ; db n,s,f[N][N]; int main() { scanf("%lf%lf"…