Find them, Catch them(POJ 1703 关系并查集)

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 38668		Accepted: 11905

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.

看食物链那一题看了半天，对并查集的理解还是不够啊！这个取余的方法有点6

#include <stdio.h>
#include <string.h>
const int maxn = 100005;

int n,m;
int p[maxn];//保存父节点
int r[maxn];//保存与父节点的关系，0为同类。1为不同类

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

int find(int x)
{
    if(x == p[x])
        return x;
    int tmp = p[x];            //记录父亲节点，方便更新r[].
    p[x] = find(p[x]);
    r[x] = (r[x]+r[tmp])%2;        //根据子节点与父亲节点的关系和父节点与爷爷节点的关系，推导子节点与爷爷节点的关系  
    return p[x];
}

void merge(int a, int b)
{
    int fa = find(a);
    int fb = find(b);
    p[fa] = fb;
    r[fa] = (r[a]+r[b]+1)%2;        //fx与x关系 + x与y的关系 + y与fy的关系 = fx与fy的关系 
}

int main()
{
    int test,a,b;
    char str[2];
    freopen("in.txt","r",stdin);
    scanf("%d",&test);
    while(test--)
    {
        scanf("%d %d",&n,&m);
        make_set();
        while(m--)
        {
            scanf("%s %d %d",str,&a,&b);

            if(str[0] == 'A')
            {
                if(find(a) != find(b))        //如果根节点不同，则不能判断关系  
                {
                    printf("Not sure yet.
");
                    continue;
                }
                else
                {
                    if(r[a] == r[b])
                        printf("In the same gang.
");
                    else printf("In different gangs.
");
                }
            }
            else
            {
                merge(a,b);
            }
        }
    }
    return 0;
}