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...
題意:給出n個構成一個環(huán)的點以及點跟點的連接關系。要求邊不能相交。比如：有一條邊i會跟j看成直線的話在園內(nèi)相交，那么可以
把一條通過邊彎繞過圓，使得兩條邊不相交。最后問是否所有線段都不相交。
分析：對于兩條邊，如果在園內(nèi)相交，則必須引導一條通過園外。把邊i看成點i1和i2分別表示通過園內(nèi)和園外。
如果i與j相交于園內(nèi)，則i1->j2，i2->j1, j2->i1, j1->i2.表示兩條邊一條在園外，一條在園內(nèi)。通過求scc確定每條邊的i1和i2
所在的scc是否相同，相同表示i既是在園內(nèi)也是在園外，矛盾了。
代碼：


#include <stdio.h>
#include <stdlib.h>
#define Min(a, b) a < b ? a : b
#define maxn 5010
struct L
{
    int x, y;
}line[500000];
struct g
{
    int v, next;
}fn[maxn * 4];
int g[maxn * 5], visit[maxn * 5], low[maxn * 5];
void set(int m)
{
    for (int i = 1; i <= m; i++)
    {
        g[i] = -1, visit[i] = 0;
    }
}
int tarjan(int u, int f, int times)
{
    low[u] = times;
    visit[u] = 1;
    int i, v;
    for (i = g[u]; i != -1; i = fn[i].next)
    {
        v = fn[i].v;
        if (v != f)
        {
            if (!visit[v])
            {
                times = tarjan(v, u, times + 1);
            }
            low[u] = Min(low[u], low[v]);
        }
    }
    return times;
}
int main()
{
    int n, m, i, j, th, times;
    scanf("%d%d", &n, &m);
        set(m * 2);
        for (i = 1; i <= m; i++)
        {
            scanf("%d%d", &line[i].x, &line[i].y);
            if (line[i].x > line[i].y)
            {
                line[i].x ^= line[i].y, line[i].y ^= line[i].x, line[i].x ^= line[i].y;
            }
        }
        for (i = 1, th = 0; i < m; i++)
        {
            for (j = i + 1; j <= m; j++)
            {
                //對相交的線段建圖 
                if (line[i].x < line[j].y && line[i].x > line[j].x && line[i].y > line[j].y
                    || line[i].y < line[j].y && line[i].y > line[j].x && line[i].x < line[j].x)
                {
                    //printf("%d %d %d %d\n", line[i].x, line[i].y, line[j].x, line[j].y);
                    fn[th].v = j + m, fn[th].next = g[i], g[i] = th++;
                    fn[th].v = i, fn[th].next = g[j+m], g[j+m] = th++;
                    fn[th].v = j, fn[th].next = g[i+m], g[i+m] = th++;
                    fn[th].v = i + m, fn[th].next = g[j], g[j] = th++;
                }
            }
        }//system("pause");
        for (i = 1, times = 1; i <= 2 * m; i++)
        {
            if (!visit[i])
            {
                times = tarjan(i, -1, times + 1);
            }
            //printf("[%d]=%d ", i, low[i]);
        }//printf("\n");
        for (i = 1; i <= m; i++)
        {
            if (low[i] == low[i+m])
            {
                break;
            }
        }
        if (i == m + 1)
        {
            printf("panda is telling the truth\n");
        }
        else
        {
            printf("the evil panda is lying again\n");
        }
    system("pause");
    return 0;
}
/*
10 3
1 5
2 6
7 3
*/