题目大意:有N串钥匙,M对锁.每串钥匙仅仅能选择当中一把.怎样选择,才干使开的锁达到最大(锁仅仅能按顺序一对一对开.仅仅要开了当中一个锁就可以) 解题思路:这题跟HDU - 3715 Go Deeper 这题的限制比較简单.都是二选一,2-SAT的裸题,仅仅只是加了二分而已 附上HDU - 3715 Go Deeper题解 #include <cstdio> #include <cstring> #include <algorithm> #include <vec…

Get Luffy Out * Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 570    Accepted Submission(s): 225 Problem Description Ratish is a young man who always dreams of being a hero. One day his friend…

二分答案 + 2-SAT验证 #include<cstdio> #include<cstring> #include<cmath> #include<stack> #include<queue> #include<algorithm> using namespace std; +; int N,M; stack<int>S; vector<int>G[maxn]; vector<int>FG[max…

HDU 1816, POJ 2723 Get Luffy Out pid=1816" target="_blank" style="">题目链接 题意:N串钥匙.每串2把,仅仅能选一把.然后有n个大门,每一个门有两个锁,开了一个就能通过,问选一些钥匙,最多能通过多少个门 思路:二分通过个数.然后对于钥匙建边至少一个不选,门建边至少一个选,然后2-sat搞一下就可以. 一開始是按每串钥匙为1个结点,但是后面发现数据有可能一把钥匙,出如今不同串(真是不合…

A Task Process Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1332    Accepted Submission(s): 656 Problem Description There are two kinds of tasks, namely A and B. There are N workers and the…

Bomb Game Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5396    Accepted Submission(s): 1925 Problem Description Robbie is playing an interesting computer game. The game field is an unbounde…

题意: 英语限制了我的行动力....就是两个钥匙不能同时用,两个锁至少开一个 建个图 二分就好了...emm....dfs  开头low 写成sccno  然后生活失去希望... #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include <set>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1816 思路:首先将每把钥匙i拆成两个点i和i+2n,分别表示选与不选,对于被分成n对的钥匙,由于只能选择其中的一把,所以加边(i,j+2n),(j,i+2n)对于每道门所对应的两把钥匙,两边中选一把,当i不选时,则必须选择j.反之,同理.所以加边(i+2n,j),(j+2n,i).二分所能打开门的数量,再用2-sat来判断可行性 . http://paste.ubuntu.com/5976255/…

题目:http://www.lydsy.com/JudgeOnline/problem.php?id=1816 分析: 我先以为是道水题,但是要注意的是每套牌中Joker只能用1张的,所以就出现了可能目前每种牌的剩余牌数都够,但不一定不用Joker,然后就短路了…… 看了hzwer的blog顿时茅塞顿开,原来是二分…… 二分答案x,然后判定 判定的方法:注意每套牌顶多只有一个Joker,所以对于答案x,能用的Joker的最大数量是T=Min(x,m),然后枚举每种牌,将每种牌相对于答案x差的个数…

Quoit Design Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 47104    Accepted Submission(s): 12318 Problem Description Have you ever played quoit in a playground? Quoit is a game in which fla…