POJ 1703 Find them, Catch them_不见不散的结局是曲终人散_新浪博客

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36652		Accepted: 11242

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：有两个帮派，其中A X Y代表判断X,Y是否为一个帮派，如果不确定就输出Not sure yet.  D,X,Y代表X，Y不属于同一个帮派；如果X,Y输入同一个帮派就输出In the same gang.否则输出In the same gang.例如：其中第一行表示输入的次数，第二行第一次数代表人的个数，第二行代表下面输入的行数；

思路：带权值的并查集，rank[]记录权值，相同的帮派间的劝值为零，不相同的帮派间的权值为一，如果D 1 2，D 1 3 怎说明2 3 属于同一个帮派；

代码如下：

#include< cstdio >

#include< cstring >

int f[100005],r[100005];

int find(int x)

{

    int tmp;

    if(x==f[x])//相等则为根节点；

        return x;

     tmp=f[x];

    f[x]=find(f[x]);

    r[x]=(r[x]+r[tmp])%2;//除以2是为了保证得到的权值为0，1两个数；

    return f[x];

   

}

int  merge(int a,int b)

{

    int fa=find(a);

    int fb=find(b);

    if(fa==fb)   return 0;//如果根节点相同说明之前已经输入过a,b,怎不用判断，直接跳过 ；

    f[fa]=fb;

    r[fa]=(-r[a]+r[b]+1)%2;//关于劝值，建议最好用手模拟一下，就容易理解了；

}

int main()

{

    int test,i,j,n,m,a,b;

    char str[2];

    scanf("%d",&test);

    while(test--)

    {

        scanf("%d%d",&n,&m);

        for(i=0; i < n;i++)

        {

            f[i]=i;

            r[i]=0;

        }

        for(i=0; i < m;i++)

        {

            scanf("%s%d%d",str,&a,&b);

            if(str[0]=='A')

            {

                if(find(a)!=find(b))

                {

                    printf("Not sure yet.\n");

                    continue;

                }