HDU 4115 Eliminate the Conflict pid=4115">题目链接 题意:Alice和Bob这对狗男女在玩剪刀石头布.已知Bob每轮要出什么,然后Bob给Alice一些限制,1表示i轮和j轮Alice必须出不一样的,0表示必须出一样的.假设Alice有一局输了就算输了,否则就是赢,问Alice能否赢 思路:2-sat问题,已经Bob出什么,Alice要么就出赢的要么就出平的,然后加上m个约束就是2-sat问题了 代码: #include <cstdio>…

Eliminate the Conflict Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1315    Accepted Submission(s): 563 Problem Description Conflicts are everywhere in the world, from the young to the elder…

Problem Description Conflicts are everywhere in the world, from the young to the elderly, from families to countries. Conflicts cause quarrels, fights or even wars. How wonderful the world will be if all conflicts can be eliminated.Edward contributes…

2-SAT,拆成六个点. #include<cstdio> #include<cstring> #include<cmath> #include<stack> #include<queue> #include<algorithm> using namespace std; +; int T,N,M; stack<int>S; vector<int>G[maxn]; vector<int>FG[max…

Problem Description Conflicts are everywhere in the world, from the young to the elderly, from families to countries. Conflicts cause quarrels, fights or even wars. How wonderful the world will be if all conflicts can be eliminated. Edward contribute…

HDU 4041 Eliminate Witches! (模拟题 ACM ICPC 2011 亚洲北京赛区网络赛题目) Eliminate Witches! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 863    Accepted Submission(s): 342 Problem Description Kaname Mado…

Eliminate the Conflict Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1114 Accepted Submission(s): 468 Problem Description Conflicts are everywhere in the world, from the young to the elderly, fr…

题意: 石头剪刀布 分别为1.2.3,有n轮,给出了小A这n轮出什么,然后m行,每行三个数a b k,如果k为0 表示小B必须在第a轮和第b轮的策略一样,如果k为1 表示小B在第a轮和第b轮的策略不一样,如果又一轮小B输了,那整个就输了,求小B能否战胜小A 解析: 写出来矛盾的情况  建图就好啦 可能我建的麻烦了...不过..我喜欢 hhhhh #include <iostream> #include <cstdio> #include <sstream> #inclu…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4115 题意:Alice和Bob玩猜拳游戏,Alice知道Bob每次会出什么,为了游戏公平,Bob对Alice的出法做出了一定限制,限制为Alice的第 i 次和第 j 次的出法相同或者不同.在n轮游戏汇总,如果Alice输了一次,那么Alice是loser. Alice每次只有两种选择,要么赢,要么平局,建立2sat模型,然后分情况建立边,分Bob第 i 次和第 j 次的拳不相等和相等两种,然后在这…

题意: 给一个字符串,表示一颗树,要求你把它整理出来,节点从1开始编号,还要输出树边. 解法: 模拟即可.因为由括号,所以可以递归地求,用map存对应关系,np存ind->name的映射,每进入一层括号,使father = now, 遇到右括号')',则father = fa[father],用vector存每个节点的子节点,然后最后dfs输出即可. 代码: #include <iostream> #include <cstdio> #include <cstring&…