ACM: 并查集 poj 1703

Findthem, Catch them

Description

The police office inTadu City decides to say ends to the chaos, as launch actions toroot up the TWO gangs in the city, Gang Dragon and Gang Snake.However, the police first needs to identify which gang a criminalbelongs to. The present question is, given two criminals; do theybelong to a same clan? You must give your judgment based onincomplete information. (Since the gangsters are always actingsecretly.)

Assume N (N <= 10^5) criminals are currently in TaduCity, numbered from 1 to N. And of course, at least one of thembelongs to Gang Dragon, and the same for Gang Snake. You will begiven M (M <= 10^5) messages in sequence, which arein the following two kinds:

1. D [a] [b]
where [a] and [b] are the numbers of two criminals, and they belongto different gangs.

2. A [a] [b]
where [a] and [b] are the numbers of two criminals. This requiresyou to decide whether a and b belong to a same gang.

Input

The first line ofthe input contains a single integer T (1 <= T<= 20), the number of test cases. Then T casesfollow. Each test case begins with a line with two integers N andM, followed by M lines each containing one message as describedabove.

Output

For each message "A[a] [b]" in each case, your program should give the judgment basedon the information got before. The answers might be one of "In thesame gang.", "In different gangs." and "Not sure yet."

Sample Input

1
5 5
A 1 2
D 1 2
A 1 2
D 2 4
A 1 4

Sample Output

Not sure yet.
In different gangs.
In the same gang.

题意: police要分别2个犯罪集团的人, 有2种操作A[a][b]:判断[a],[b]是否在同一个集团里面.

      D[a][b]:将[a],[b]分到2个集团中去.

 

解题思路:

    1. 明显的并查集问题, 这题有一个小问题就是D操作时分配到不同的集合, 因此每一个人对咬设置

       多一个作为他本身的"反对称".

       即是: [a]的反对称设置为[a]+n; 当每次分配的时候D[a][b]时:

             union_set(a,b+n), union_set(b, a+n); 设置的"反对称"可以作为一个标记, 当"反对称"

             b+n与a同时出现时, 可以判定[a],[b]是不同集团的.

代码:

#include<cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 100005

int n, m;
int p[MAX*2];
int a, b;
char ch[2];

void init()
{
 for(int i = 0; i <= 2*n;++i)
  p[i] = i;
}

int find(int x)
{
 return p[x] == x ? x : (p[x] = find(p[x]));
}

inline void union_set(int x, int y)
{
 x = find(x);
 y = find(y);
 if(x != y) p[x] = y;
}

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

  for(int i = 0; i< m; ++i)
  {
   scanf("%s %d%d",ch, &a, &b);
   if(ch[0] =='D')
   {
    union_set(a,n+b);
    union_set(b,n+a);
   }
   else if(ch[0]== 'A')
   {
    if(find(a)== find(b))
     printf("Inthe same gang.\n");
    elseif(find(a) == find(b+n) || find(b) == find(a+n))
     printf("Indifferent gangs.\n");
    else
     printf("Notsure yet.\n");
   }
  }
 }
 return 0;
}