poj 1703 Find them, Catch them（种类并查集和一种巧妙的方法）

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36176		Accepted: 11090

Description

The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon and Gang Snake. However, the police first needs to identify which gang a criminal belongs to. The present question is, given two criminals; do they belong to a same clan? You must give your judgment based on incomplete information. (Since the gangsters are always acting secretly.) 

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

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

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

Input

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

Output

For each message "A [a] [b]" in each case, your program should give the judgment based on the information got before. The answers might be one of "In the same 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：是一个城市有两个帮派，A 1 2是询问1和2这两个人是不是一个帮派，D 1 2是这两个人不在一个帮派
	每一次A就输出一次
思路： 带权并查集，加一个数组re[]表示子节点和其父节点的关系，是一个门派就是0。不是就是1
2015,7,27
#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
#define M 100100
int x[M],re[M];
void init()
{
	for(int i=0;i<M;i++){
		x[i]=i;
		re[i]=0;
	}
}
int find(int k)
{
	int temp=x[k];
	if(x[k]==k) return k;
	x[k]=find(x[k]);
	if(re[k]==re[temp])//假设和父节点是同类就存0
		re[k]=0;
	else re[k]=1;
	return x[k];
}
void merge(int a,int b,int fa,int fb)
{
	x[fa]=fb;
	if(re[a]==re[b]) re[fa]=1;
	else re[fb]=0;
}
int main()
{
	int t,n,m,i,a,b,fa,fb;
	char ch;
	scanf("%d",&t);
	while(t--){
		init();
		scanf("%d%d",&n,&m);
		while(m--){
			cin>>ch;
			scanf("%d%d",&a,&b);
			fa=find(a);
			fb=find(b);
			if(ch=='A'){				
				if(fa!=fb){
					printf("Not sure yet.
");
					continue;
				}
				if(re[a]==re[b]){
					printf("In the same gang.
");
					continue;
				}
				else{
					printf("In different gangs.
"); 
				}
			}
			else{
				if(fa!=fb)
					merge(a,b,fa,fb);
			}
		}
	}	
	return 0;
} 
/* 还有一种巧妙方法：大神的解析。是poj食物链那道题的弱化。
 
可见全部元素个数为2 * N，假设i表示属于帮派A。那么i + N表示属于帮派B。
每次输入两个家伙不在同一帮派的时候。就合并他们分属两个帮派的元素。
*/
#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
#define M 200200
int x[M];
void init()
{
	for(int i=0;i<M;i++)
		x[i]=i;
}
int find(int k)
{
	if(x[k]==k) return k;
	x[k]=find(x[k]);//压缩路径 
	return x[k];
}
void merge(int a,int b)
{
	int fa=find(a);
	int fb=find(b);
	if(x[a]!=x[b]) x[fa]=fb;
}
bool same(int a,int b)
{
	return find(a)==find(b);
}
int main()
{
	int t,n,m,i,a,b,fa,fb;
	char ch;
	scanf("%d",&t);
	while(t--){
		init();
		scanf("%d%d",&n,&m);
		while(m--){
			cin>>ch;
			scanf("%d%d",&a,&b);
			if(ch=='A'){				
				if(same(a,b)){
					printf("In the same gang.
");
					continue;
				}
				if(same(a,b+n)){
					printf("In different gangs.
"); 
				}
				else{
					printf("Not sure yet.
");
					continue;
				}
			}
			else{
				merge(a,b+n);//a和b一定不在一个门派，就把a和b+n合并 
				merge(a+n,b);//同理 
			}
		}
	}	
	return 0;
}