原文地址：A Bug's Life -Poj作者：蘭多夫87
A Bug's Life
Time Limit: 10000MS Memory Limit:
65536K
Total Submissions: 25599 Accepted:
8332
Description
Background
Professor Hopper is researching the sexual behavior of a rare
species of bugs. He assumes that they feature two different genders
and that they only interact with bugs of the opposite gender. In
his experiment, individual bugs and their interactions were easy to
identify, because numbers were printed on their
backs.
Problem
Given a list of bug interactions, decide whether the
experiment supports his assumption of two genders with no
homosexual bugs or if it contains some bug interactions that
falsify it.
Input
The first line of the input contains the number of scenarios.
Each scenario starts with one line giving the number of bugs (at
least one, and up to 2000) and the number of interactions (up to
1000000) separated by a single space. In the following lines, each
interaction is given in the form of two distinct bug numbers
separated by a single space. Bugs are numbered consecutively
starting from one.
Output
The output for every scenario is a line containing "Scenario
#i:", where i is the number of the scenario starting at 1, followed
by one line saying either "No suspicious bugs found!" if the
experiment is consistent with his assumption about the bugs' sexual
behavior, or "Suspicious bugs found!" if Professor Hopper's
assumption is definitely wrong.
Sample
Input
2
3 3
1 2
2 3
1 3
4 2
1 2
3 4
Sample
Output
Scenario #1:
Suspicious bugs found!
Scenario #2:
No suspicious bugs found!
Hint
Huge input,scanf is recommended.
Source
TUD Programming Contest 2005, Darmstadt, Germany
主要算法：並查集
我們用f[i]表示i的父親，用v[i]表示i與他父親的關係，0表示同性，1表示異性
首先我們初始化，f[i]=i;v[i]=0;
每次給出兩隻蟲子，我們先判斷他們的祖先是否相同，如果相同，那他們的關係我們是已知的，可以通過他們與祖先的關係推出，再判斷是否矛盾
如果不相同，就把它們的祖先的關係推出來，再把他們的祖先建立關係
注意路徑壓縮時，v[x]一起更新，不要弄錯了
代碼：