POJ 1703 Find them, Catch them

H - Find them, Catch them

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

Submit

Status

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.

看了三天，又百度了下，，，，，

思路：

两种种族并查集在链表里是以ABABABABAB连接的，所以只需求出它与根的距离。

合并也要注意，

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MA 100100
int fer[MA];
int lei[MA];
int find(int xx)
{
int p=xx;
lei[p]=0;
while (fer[xx]!=xx)
{
lei[p]++;//求距离
xx=fer[xx];
}
if (lei[p]%2==1)
fer[p]=xx;
else
fer[fer[p]]=xx;
return xx;
}
void uion(int aa,int bb)
{
int fa=find(aa);
int fb=find(bb);
if (fa==fb) return ;//避免死循环。
if ((lei[aa]%2)==(lei[bb]%2))//开始合并。
fer[fa]=fb;
else if ((lei[aa]%2)==0)
fer[fa]=bb;
else
fer[fb]=aa;
}
int main()
{
int t;
scanf("%d",&t);
while (t--)
{
int n,m,a,b;
char ch[5];
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
fer[i]=i;
for (int i=0;i<m;i++)
{
scanf("%s%d%d",&ch,&a,&b);
if (ch[0]=='A')
{
if (find(a)==find(b))
{
if ((lei[a]%2)==(lei[b]%2))
printf("In the same gang.\n");
else
printf("In different gangs.\n");
}
else
{
if(n==2) printf("In different gangs.\n");
else printf("Not sure yet.\n");
}
}
else
{
uion(a,b);
}
}
}
return 0;
}