Tempter of the Bone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 110290    Accepted Submission(s): 29967 Problem Description The doggie found a bone in an ancient maze, which fascinated him a…

思路: 剪枝的思路参考博客:http://www.cnblogs.com/zibuyu/archive/2012/08/17/2644396.html  在其基础之上有所改进 题意可以给抽象成给出一个图,让你求S点到D点之间是否存在一条长度为T的道路.求两地之间的距离用的是dfs,而dfs在这里的关键是找到回溯的条件,就是当到达D点并且剩余步数为0时,则符合题意的要求,由于我们只需要知道这样一条长度为T的路径是否存在,因此当我们发现存在的时候,只需要将一个全局flag给设置为1即可,然后从此之后…

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…

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…

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…

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.…

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;…

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…

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…

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…