POJ 3207：Ikki's Story IV

Ikki's Story IV - Panda's Trick

Time Limit: 1000MS Memory Limit: 131072KTotal Submissions: 10073 Accepted: 3703

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 20 13 2

Sample Output

panda is telling the truth...

题意：

一个圆边上有n个点（标号顺时针），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。输入保证每个点最多只会连接的一条边。问有没有方法连接这m条边，使这些边都不相交

建图方式参考

http://blog.csdn.net/jaihk662/article/details/71087449

直接强联通分量，若某个强联通分量中的一个点要选，那么这个强联通分量中的所有点都必选

如果某个点对中的两个点属于同一强联通分量，那么无解，否则一定有解

复杂度O(n)

#include<stdio.h>#include<string.h>#include<algorithm>#include<stack>#include<vector>using namespace std;vector<int> G[2017];stack<int> st;typedef struct{int x;int y;}Res;Res s[1005];int t, num, time[1005], low[1005], vis[1005], fa[1005];void tarjan(int u){int v, i;vis[u] = 1;st.push(u);time[u] = low[u] = ++t;for(i=0;i<G[u].size();i++){v = G[u][i];if(time[v]==0){tarjan(v);low[u] = min(low[u], low[v]);}else if(vis[v])low[u] = min(low[u], time[v]);}if(time[u]==low[u]){num++;while(1){  v = st.top();st.pop();vis[v] = 0;fa[v] = num;if(v==u)break;}}}int main(void)  {int i, j, ans, n;scanf("%*d%d", &n);for(i=1;i<=n;i++){scanf("%d%d", &s[i].x, &s[i].y);s[i].x++; s[i].y++;if(s[i].x>s[i].y)swap(s[i].x, s[i].y);}for(i=1;i<=n*2;i++)G[i].clear();for(i=1;i<=n;i++){for(j=i+1;j<=n;j++){if((s[i].x<=s[j].x && s[i].y>=s[j].x && s[i].y<=s[j].y) || (s[i].x>=s[j].x && s[i].x<=s[j].y && s[i].y>=s[j].y)){G[i].push_back(j+n);G[j].push_back(i+n);G[i+n].push_back(j);G[j+n].push_back(i);}}}n = 2*n;  for(i=1;i<=n;i++){if(time[i]==0)tarjan(i);}ans = 1;for(i=1;i<=n/2;i++){        if(fa[i]==fa[i+n/2])ans = 0;}if(ans)printf("panda is telling the truth...\n");elseprintf("the evil panda is lying again\n");return 0;}