图论--2-SAT--POJ Ikki's Story IV - Panda's Trick

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 ai and bi, 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...

题意

平面上，一个圆，圆的边上按顺时针放着n个点。现在要连m条边，比如a，b，那么a到b可以从圆的内部连接，也可以从圆的外部连接。给你的信息中，每个点最多只会连接的一条边。问能不能连接这m条边，使这些边都不相交。

分析：

如1 5连边，2，6连边，由于点是顺序排列的，一画图就可以发现，这两条边必须一个从圆外面连，一个从内部连，否则就会相交。如果再加入3 7这条边，那么就必须相交了。

这样，就可以转化成标准的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。计作点i和点i'

2：如果两条边i和j必须一个画在内部，一个画在外部（一个简单判断就可以）

建图：看我博客2-SAT详解,比较水的建图方式，就不再这写了。

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <algorithm>
#define MAXN 4000
#define MAXM 500000
#define INF 10000000
using namespace std;
int n, m;
struct Edge
{
	int s, t;//起点 终点
}num[MAXM];//存储每条线
vector<int> G[MAXN];
int low[MAXN], dfn[MAXN];
int dfs_clock;
int sccno[MAXN], scc_cnt;
stack<int> S;
bool Instack[MAXN];
void init()
{
	for(int i = 1; i <= 3*m; i++) G[i].clear();
}
bool judge(int a, int b, int c, int d)//判断是否相交
{
	return (c > a && b > c && d > b) || (a > c && d > a && b > d);
}
void getMap()
{
	for(int i = 1; i <= m; i++) scanf("%d%d", &num[i].s, &num[i].t);
	/* i + n表内 i +2*n表外*/
	for(int i = 1; i <= m; i++)
	{
		for(int j = 1; j < i; j++)
		{
			if(judge(num[i].s, num[i].t, num[j].s, num[j].t))//相交
			{
				G[i + 2*m].push_back(j + m);//i外 -> j内
				G[i + m].push_back(j + 2*m);//i内 -> j外
				G[j + m].push_back(i + 2*m);//j内 -> i外
				G[j + 2*m].push_back(i + m);//j外 -> i内
			}
		}
	}
}
void tarjan(int u, int fa)
{
	int v;
	low[u] = dfn[u] = ++dfs_clock;
	S.push(u);
	Instack[u] = true;
	for(int i = 0 ;i < G[u].size(); i++)
	{
		v = G[u][i];
		if(!dfn[v])
		{
			tarjan(v, u);
			low[u] = min(low[v], low[u]);
		}
		else if(Instack[v])
		low[u] = min(low[u], dfn[v]);
	}
	if(low[u] == dfn[u])
	{
		scc_cnt++;
		for(;;)
		{
			v = S.top(); S.pop();
			Instack[v] = false;
			sccno[v] = scc_cnt;
			if(v == u) break;
		}
	}
}
void find_cut(int l, int r)//tarjan求SCC
{
	memset(low, 0, sizeof(low));
	memset(dfn, 0, sizeof(dfn));
	memset(sccno, 0, sizeof(sccno));
	memset(Instack, false, sizeof(Instack));
	dfs_clock = scc_cnt = 0;
	for(int i = l; i <= r; i++)
	if(!dfn[i]) tarjan(i, -1);
}
void solve()
{
	bool flag = true;
	for(int i = 1; i <= m; i++)
	{
		if(sccno[i + m] == sccno[i + 2*m])
		{
			flag = false;
			printf("the evil panda is lying again\n");
			break;
		}
	}
	if(flag)
	printf("panda is telling the truth...\n");
}
int main()
{
	scanf("%d%d", &n, &m);
	init();
	getMap();
	find_cut(1, 3*m);//按边处理  总编号为3*m
	solve();
	return 0;
}