POJ 3905 Perfect Election id=3905" target="_blank" style="">题目链接 思路:非常裸的2-sat,就依据题意建边就可以 代码: #include <cstdio> #include <cstring> #include <cstdlib> #include <vector> #include <algorithm> using nam…

Perfect Election Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 438   Accepted: 223 Description In a country (my memory fails to say which), the candidates {1, 2 ..., N} are running in the parliamentary election. An opinion poll asks th…

题意:有N个人参加选举,有M个条件,每个条件给出:i和j竞选与否会只要满足二者中的一项即可.问有没有方案使M个条件都满足. 分析:读懂题目即可发现是2-SAT的问题.因为只要每个条件中满足2个中的一个即可,因此将人i拆成 点i表示不竞选,和点i+N表示竞选,根据合取式的满足条件建图跑Tarjan. 最后检查每个i与i+N是否在同一强连通分量中. #include<stdio.h> #include<iostream> #include<algorithm> #inclu…

2-SAT 裸题,搞之 #include<cstdio> #include<cstring> #include<cmath> #include<stack> #include<vector> #include<algorithm> using namespace std; +; int N,M; ],s2[]; stack<int>S; vector<int>G[maxn]; vector<int>…

Perfect Election Time Limit: 5000MS         Memory Limit: 65536K Total Submissions: 964         Accepted: 431 Description In a country (my memory fails to say which), the candidates {1, 2 ..., N} are running in the parliamentary election. An opinion…

POJ 3398 Perfect Service(树型动态规划,最小支配集) Description A network is composed of N computers connected by N − 1 communication links such that any two computers can be communicated via a unique route. Two computers are said to be adjacent if there is a com…

1.链接地址: http://bailian.openjudge.cn/practice/2810/ http://bailian.openjudge.cn/practice/1543/ http://poj.org/problem?id=1543 2.题目: Perfect Cubes Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 13190   Accepted: 6995 Description For hundr…

Description Given a connected undirected graph, tell if its minimum spanning tree is unique.  Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V', E'), with the followi…

题目链接: http://poj.org/problem?id=1679 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G…

LINK 题意:询问是否存在直线,使得所有线段在其上的投影拥有公共点 思路:如果投影拥有公共区域,那么从投影的公共区域作垂线,显然能够与所有线段相交,那么题目转换为询问是否存在直线与所有线段相交.判断相交先求叉积再用跨立实验.枚举每个线段的起始结束点作为直线起点终点遍历即可. /** @Date : 2017-07-12 14:35:44 * @FileName: POJ 3304 基础线段交判断.cpp * @Platform: Windows * @Author : Lweleth (Sou…