题意: 圆上n个点,m对点之间连边,连在园内或园外,所有边不相交是否可行 发现两对点连线都在内相交则都在外也相交,那么只有一个在内一个在外啦,转化为$2-SAT$问题 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; ,M=5e5; typedef long long…

Ikki's Story IV - Panda's Trick Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 7296   Accepted: 2705 Description liympanda, one of Ikki’s friend, likes playing games with Ikki. Today after minesweeping with Ikki and winning so many tim…

http://poj.org/problem?id=3207 Ikki's Story IV - Panda's Trick Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 7021   Accepted: 2604 Description liympanda, one of Ikki’s friend, likes playing games with Ikki. Today after minesweeping wi…

POJ 3207 Ikki's Story IV - Panda's Trick(2-sat问题) Description liympanda, one of Ikki's friend, likes playing games with Ikki. Today after minesweeping with Ikki and winning so many times, he is tired of such easy games and wants to play another game…

Ikki's Story IV - Panda's Trick Time Limit: 1000MS   Memory Limit: 131072K Total Submissions: 6691   Accepted: 2496 Description liympanda, one of Ikki’s friend, likes playing games with Ikki. Today after minesweeping with Ikki and winning so many tim…

POJ 3207 Ikki's Story IV - Panda's Trick liympanda, one of Ikki's friend, likes playing games with Ikki. Today after minesweeping with Ikki and winning so many times, he is tired of such easy games and wants to play another game with Ikki. liympanda…

http://poj.org/problem?id=3207 题意:一个圆上顺时针依次排列着标号为1-n的点,这些点之间共有m条边相连,每两个点只能在圆内或者圆外连边.问是否存在这些边不相交的方案.(n<=1000, m<=500) #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <iostream> using n…

题意: 有一个环,环上n个点,现在在m个点对之间连一条线,线可以往圆外面绕,也可以往里面绕,问是否必定会相交? 思路: 根据所给的m条边可知,假设给的是a-b,那么a-b要么得绕环外,要么只能在环内,除非a和b是连续的点才不会影响到任何弧,否则一定会多少影响其他弧的走势.比如样例所举出的 0-2和 1-3,就必须有一条弧选择外环,一个选择内环. 如何使他们有序不冲突呢?其实这题跟环上的点没多大关系,而且给的点数n也没什么用,而倒是边才重要.我们要做的是让边不冲突.那么肯定是要先选出可能冲突的边来…

题意: 就是一个圈上有n个点,给出m对个点,这m对个点,每一对都有一条边,合理安排这些边在圈内或圈外,能否不相交 解析: 我手残 我手残 我手残 写一下情况 只能是一个在圈外 一个在圈内 即一个1一个0 #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <cctype> #include…

注意到相交的点对一定要一里一外,这样就变成了2-SAT模型 然后我建边的时候石乐志,实际上不需要考虑这个点对的边是正着连还是反着连,因为不管怎么连,能相交的总会相交,所以直接判相交即可 然后tarjan缩点,再判是否合法即可 #include<iostream> #include<cstdio> using namespace std; const int N=100005; int n,m,h[N],cnt,l[N],r[N],dfn[N],low[N],tot,s[N],top…