2-sat->poj 3207 Ikki's Story IV - Panda's Trick

最新推荐文章于 2019-08-13 17:32:00 发布

原创最新推荐文章于 2019-08-13 17:32:00 发布 · 537 阅读

0 ·

CC 4.0 BY-SA版权

ACM->2-sat 专栏收录该内容

10 篇文章

订阅专栏

本文介绍了一个关于连线的游戏问题，玩家需要判断是否能在圆内或圆外连接指定的点对而不让连线在圆内或圆外交叉（除了边界）。通过构建图论模型并使用Tarjan算法来解决该问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Ikki's Story IV - Panda's Trick

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 7258		Accepted: 2688

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 with Ikki.

liympanda has a magic circle and he puts it on a plane, there are n points on its boundary in circular border: 0, 1, 2, …, n − 1. Evil panda claims that he is connecting m pairs of points. To connect two points, liympanda either places the link entirely inside the circle or entirely outside the circle. Now liympanda tells Ikki no two links touch inside/outside the circle, except on the boundary. He wants Ikki to figure out whether this is possible…

Despaired at the minesweeping game just played, Ikki is totally at a loss, so he decides to write a program to help him.

Input

The input contains exactly one test case.

In the test case there will be a line consisting of of two integers: n and m (n ≤ 1,000, m ≤ 500). The following m lines each contain two integers a_i and b_i, which denote the endpoints of the ith wire. Every point will have at most one link.

Output

Output a line, either “panda is telling the truth...” or “the evil panda is lying again”.

Sample Input

4 2
0 1
3 2

Sample Output

panda is telling the truth...

2-sta问题：

1：每个边看成2个点：分别表示在内部连接和在外部连接，只能选择一个。计作i和i'

2：如果两条边i和j必须一个画在内部，一个画在外部

连边：

i->j’,

j->i’，

i’->j，

j’->i，

#include <iostream>
#include <cstdio>
#include <string.h>
#include <stack>

using namespace std;

#define MAXN 500 + 10
#define MAXV 1005

int c[MAXN][2];
int n, m;

struct Edge
{
	int v, next;
}edge[MAXV * MAXV];

int head[2 * MAXN], e, cnt;
int dfn[2 * MAXN], low[2 * MAXN], belong[2 * MAXN], ind;
bool ins[2 * MAXN];

void add(int u, int v)
{
	edge[e].v = v;
	edge[e].next = head[u];
	head[u] = e++;
}

void init()
{
	e = cnt = ind = 0;
	memset(head, -1, sizeof(head));
}

stack <int> s;

void tarjan(int u)
{
	dfn[u] = low[u] = ++ind;
	
	ins[u] = true;
	s.push(u);
	
	int v;
	
	for (int i = head[u]; i != -1; i = edge[i].next)
	{
		v = edge[i].v;
		if (!dfn[v])
		{
			tarjan(v);
			
			low[u] = min(low[u], low[v]);
		}
		else if (ins[v])
		{
			low[u] = min(low[u], dfn[v]);
		}
	}
	
	if (low[u] == dfn[u])
	{
		cnt++;
		
		int x;
		
		do
		{
			x = s.top();
			s.pop();
			ins[x] = false;
			belong[x] = cnt;
		}while (x != u);
	}
}

bool solve()
{
	for (int i = 0; i < m; i++)
	{
		if (!dfn[i])
		{
			tarjan(i);
		}
	}
	
	for (int i = 0; i < m; i++)
	{
		if (belong[i] == belong[i + m])
		{
			return false;
		}
	}
	
	return true;
}

void input()
{
	init();
	
	cin >> n >> m;
	
	for (int i = 0; i < m; i++)
	{
		cin >> c[i][0] >> c[i][1];
		if (c[i][0] > c[i][1])
		{
			swap(c[i][0], c[i][1]);
		}
	}
	
	for (int i = 0; i < m; i++)
	{
		for (int j = i + 1; j < m; j++)
		{
			if (c[i][0] > c[j][0] && c[i][0] < c[j][1] && c[i][1] > c[j][1] ||
				c[i][1] > c[j][0] && c[i][1] < c[j][1] && c[i][0] < c[j][0])
			{
				add(i, j + m);
				add(j, i + m);
				add(i + m, j);
				add(j + m, i);			
			}
		}
	}
	
	if (solve())
	{
		cout << "panda is telling the truth..."	 << endl;
	}
	else
	{
		cout << "the evil panda is lying again" << endl;
	}
}

int main()
{
	input();
	return 0;
}