UVa 208 消防车（dfs+剪枝）

Sticks(UVA - 307)【DFS+剪枝】

Sticks(UVA - 307) 题目链接算法 DFS+剪枝 1.这道题题意就是说原本有一些等长的木棍,后来把它们切割,切割成一个个最长为50单位长度的小木棍,现在想让你把它们组合成一个个等长的大木棍,要求这个拼接成的大木棍的长度最小.问最小长度是多少.(注意,在接下来的介绍中,将最后的大木棍表述为拼接木棍,小木棍还是叫小木棍) 2.看完这个题,基本思路是从一个小木棍开始找,找到一个未使用的小木棍后拼接,当拼接成的长度是所有小木棍长度和的约数时,就暂时把它定义为拼接木棍的最小长度,然后计算最…

UVa 208 消防车（dfs+剪枝）

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=144 题意:给出一个n个结点的无向图以及某个结点k,按照字典序从小到大顺序输出从1到结点k的所有路径. 思路:如果直接矩阵深搜的话是会超时的,所以我们可以从终点出发,将与终点相连的连通块保存起来,这样dfs深搜时可以剪枝掉一些到达不了的点.只要解决了这个,dfs就是小问题. 这道题还有点坑的…

UVa 208 - Firetruck 回溯+剪枝数据

题意:构造出一张图,给出一个点,字典序输出所有从1到该点的路径. 裸搜会超时的题目,其实题目的数据特地设计得让图稠密但起点和终点却不相连,所以直接搜索过去会超时. 只要判断下起点和终点能不能相连就行了,可以用并查集也可以用floyd算法,这样就能过了. 但是这个方法不是很完美的,如果两点之间只有一条线相连,而图又是稠密图,这样也很容易超时,数据强电就会挂掉. 可以把算法改进一下:是先从终点出发,无回溯的走遍和终点相连的所有点并标记,然后从起点出发,DFS判断下标记,这样就不会多走很多路了.另一个…

*HDU1455 DFS剪枝

Sticks Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9779 Accepted Submission(s): 2907 Problem Description George took sticks of the same length and cut them randomly until all parts became…

POJ 3009 DFS+剪枝

POJ3009 DFS+剪枝原题: Curling 2.0 Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 16280 Accepted: 6725 Description On Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules are somewhat different from our…

poj 1724:ROADS（DFS + 剪枝）

ROADS Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10777 Accepted: 3961 Description N cities named with numbers 1 ... N are connected with one-way roads. Each road has two parameters associated with it : the road length and the toll…

DFS(剪枝) POJ 1011 Sticks

题目传送门 /* 题意:若干小木棍,是由多条相同长度的长木棍分割而成,问最小的原来长木棍的长度: DFS剪枝:剪枝搜索的好题!TLE好几次,终于剪枝完全! 剪枝主要在4和5:4 相同长度的木棍不再搜索:5 若新的搜索连第一条都没组合出来,直接break: 详细解释:http://blog.csdn.net/lyy289065406/article/details/6647960 http://www.cnblogs.com/devil-91/archive/2012/08/03/2621787.…

DFS+剪枝 HDOJ 5323 Solve this interesting problem

题目传送门 /* 题意:告诉一个区间[L,R],问根节点的n是多少 DFS+剪枝:父亲节点有四种情况:[l, r + len],[l, r + len - 1],[l - len, r],[l - len -1,r]; */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <queue> using namespace std;…

HDU 5952 Counting Cliques 【DFS+剪枝】（2016ACM/ICPC亚洲区沈阳站）

Counting Cliques Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 539 Accepted Submission(s): 204 Problem Description A clique is a complete graph, in which there is an edge between every pair…

HDU 5937 Equation 【DFS+剪枝】（2016年中国大学生程序设计竞赛（杭州））

Equation Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 92 Accepted Submission(s): 24 Problem Description Little Ruins is a studious boy, recently he learned addition operation! He was rewa…

LA 6476 Outpost Navigation （DFS+剪枝）

题目链接 Solution DFS+剪枝对于一个走过点k,如果有必要再走一次,那么一定是走过k后在k点的最大弹药数增加了.否则一定没有必要再走. 记录经过每个点的最大弹药数,对dfs进行剪枝. #include <iostream> #include <cstring> #include <algorithm> #include <cstdio> #include <map> using namespace std; map<string…

poj 1011 Sticks （DFS+剪枝）

Sticks Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 127771 Accepted: 29926 Description George took sticks of the same length and cut them randomly until all parts became at most 50 units long. Now he wants to return sticks to the or…

poj 1564 Sum It Up | zoj 1711 | hdu 1548 (dfs + 剪枝 or 判重)

Sum It Up Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Total Submission(s) : 4 Accepted Submission(s) : 1 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description Given a specified total t and…

hdu 4109 dfs+剪枝优化

求最久时间即在无环有向图里求最远路径 dfs+剪枝优化从0节点(自己添加的)出发,0到1~n个节点之间的距离为1.mt[i]表示从0点到第i个节点眼下所得的最长路径 #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<vector> using namespace std; const…

PKU 2531 Network Saboteur(dfs+剪枝||随机化算法)

题目大意:原题链接给定n个节点,任意两个节点之间有权值,把这n个节点分成A,B两个集合,使得A集合中的每一节点与B集合中的每一节点两两结合(即有|A|*|B|种结合方式)权值之和最大. 标记:A集合:true B集合:false 解法一:dfs+剪枝 #include<iostream> #include<cstring> using namespace std; int n,ans; ]; ][]; void dfs(int i,int cursum) { in[i]=tru…

HDU 1501 Zipper 【DFS+剪枝】

HDU 1501 Zipper [DFS+剪枝] Problem Description Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in…

HDOJ 1501 Zipper 【DP】【DFS+剪枝】

HDOJ 1501 Zipper [DP][DFS+剪枝] Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10886 Accepted Submission(s): 3925 Problem Description Given three strings, you are to determine whether the third str…

poj 1011 ：Sticks (dfs+剪枝)

题意:给出n根小棒的长度stick[i],已知这n根小棒原本由若干根长度相同的长木棒(原棒)分解而来.求出原棒的最小可能长度. 思路:dfs+剪枝.蛮经典的题目,重点在于dfs剪枝的设计.先说先具体的实现:求出总长度sum和小棒最长的长度max,则原棒可能的长度必在max~sum之间,然后从小到大枚举max~sum之间能被sum整除的长度len,用dfs求出所有的小棒能否拼凑成这个长度,如果可以,第一个len就是答案. 下面就是关键的了,就是这道题dfs的实现和剪枝的设计: 1.以一个小棒为开头…

Black And White(DFS+剪枝)

Black And White Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 3937 Accepted Submission(s): 1082Special Judge Problem Description In mathematics, the four color theorem, or the four color…

codeforces 615 B. Longtail Hedgehog (DFS + 剪枝)

题目链接: codeforces 615 B. Longtail Hedgehog (DFS + 剪枝) 题目描述: 给定n个点m条无向边的图,设一条节点递增的链末尾节点为u,链上点的个数为P,则该链的beauty值 = P*degree[u].问你所有链中最大的beauty值. 解题思路: 因为只需要找到以每个节点为终点的最长递增链即可,所以建图的时候可以建成从编号小的节点到编号大的节点的有向图,然后用DFS搜索,但是直接暴搜会TLE,然后就开始了我的TLE之路,TLE24, TLE42,TL…