6450 Social AdvertisingYou have decided to start up a new social networking company. Other existing popular social networksalready have billions of users, so the only way to compete with them is to include novel features noother networks have.Your co…

[题目] 给一个无向图,每当对某个点操作,该点以及与该点相连的点都获得标记,问标记所有点至少需要操作多少次 输入 第一行为T,表示测试数据组数 每组测试数据第一行为n(1<=n<=20)表示有n个点,接下来n行,第i行描述与结点i所连的点,首个整数d,接下来有d个整数,表示结点i与它之间有一条边,该图没有自环 输出 每组测试数据输出一行,一个数表示至少需要操作的次数 [题解] 状态压缩+枚举 这道题数据范围好小,那么就应该不是什么特殊的算法了,可以考虑下搜索 将每个点与其他点的关系压缩,记录到…

题意:一些人有朋友关系,在某个人的社交网站上投放广告可以被所有该人的直接朋友看到,问最小投放多少个广告使给出的人都看到广告.(n<=20) 解法:看到n的范围可以想到用二进制数表示每个人被覆盖与否,所以可以依次为状态进行搜索,每次枚举一个人,投放广告,然后将他的朋友覆盖,用dis记录步数,记忆化搜索. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath>…

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…

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…