ACM: 并查集 poj 2492

A Bug's Life

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!

 

题意: Professor Hopper想试验检测虫子是否有异常的性行为(同性恋), 每个交互代表一次关系.

      最后观察是否有同性恋情况.

 

解题思路:

     1. 并查集问题, 这题要加上关系集合的处理. 用0,1来代表两个性别.

     2. 如果两只虫发生关系, 需要将两只小虫分到2个集合中即可.

 

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 2005

int n, m;
int p[MAX], rel[MAX];

void init()
{
 for(int i = 1; i <= n; ++i)
 {
  p[i] = i;
  rel[i] = 0;
 }
}

int find(int x)
{
 if(x != p[x])
 {
  int temp = p[x];
  p[x] = find(p[x]);
  rel[x] = rel[x]^rel[temp];
 }
 return p[x];
}

void union_set(int a, int b)
{
 int pa, pb;
 pa = find(a);
 pb = find(b);
 p[pb] = pa;
 rel[pb] = ~(rel[b]^rel[a]);
}

int main()
{
// freopen("input.txt", "r", stdin);
 int caseNum, num = 1;
 scanf("%d",&caseNum);
 while(caseNum--)
 {
  scanf("%d %d",&n, &m);
  init();

  int a, b;
  bool flag = false;
  for(int i = 1; i <= m; ++i)
  {
   scanf("%d %d",&a, &b);
   int pa, pb;
   pa = find(a);
   pb = find(b);
   if(pa != pb)
    union_set(a, b);
   else
   {
    if(rel[a] == rel[b]) flag = true;
   }
  }

  printf("Scenario #%d:\n", num++);
  if(flag)
   printf("Suspicious bugs found!\n\n");
  else
   printf("No suspicious bugs found!\n\n");
 }
 return 0;
}