POJ 1703 Find them, Catch them (关系并查集)

这道题目的描述我感觉还是抄啸爷的吧，这个描述太逗了啊。

题意：某城市存在蛇帮和龙帮两大帮派（话说名字还能再挫一点点吗. . . ）。在某一次像东莞这样的大规模的扫黄打非中，警察抓住了 n 个人，但是不缺定他们分别属于哪个帮派。现在给你一些条件，然后让你判断两个人是否属于同一个帮派。

这是一道关系并查集的题目啊，第一次做。啸爷很耐心的讲解了思路。过程很好理解啊。

就是这个并查集需要保存一下他自己与父亲节点的关系，0代表不是一伙，1代表是一伙的。然后在向跟回溯的时候找到这个节点和父亲节点的关系，如果相同r就 不改变，否则就改变。一直找到根节点判断节点和根节点的关系。比较一下就知道是否是一个帮派了啊。因为除了0，就是1。

啸爷的解释：

关系并查集的合并：

当不属于同一棵树的两个点 u , v确定关系时，要将两棵树进行合并。

仍以此题为例：

设ru，rv分别为u，v的跟节点，wu，wv分别表示u，v与跟节点的关系，w[]为节点与父节点的关系。

当wu == wv时，则表示ru和rv属于不同帮派（因为此题中，每次确定关系均为不属于同一帮派）, 则w[ru] = 1，fa[ru] = rv；

当wu != wv时，则表示ru和rv属于同一帮派 ， 则w[ru] = 0，fa[ru] = rv；

Find them, Catch them

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 27932 Accepted: 8521

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

15 5A 1 2D 1 2A 1 2D 2 4A 1 4

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>#define eps 1e-7#define M 10001000#define LL __int64//#define LL long long#define INF 0x3f3f3f3f#define PI 3.1415926535898const int maxn = 101000;using namespace std;int f[maxn];int w[maxn];int r;int _find(int x){    int t = x;    r = 0;    while(x != f[x])    {        if(w[x])        {            if(r == 1)                r = 0;            else                r = 1;        }        x = f[x];    }    f[t] = x;    w[t] = r;    return x;}int main(){    int t;    cin >>t;    int n, m;    while(t--)    {       scanf("%d %d",&n,&m);        char str;        for(int i = 0; i <= n; i++)        {            f[i] = i;            w[i] = 0;        }        while(m--)        {            scanf("%*c%c",&str);            int x, y;            if(str == 'A')            {                scanf("%d %d",&x, &y);                int xx = _find(x);                int r1 = r;                int yy = _find(y);                int r2 = r;                if(xx != yy)                   printf("Not sure yet.\n");                else                {                    if(r1 == r2)                       printf("In the same gang.\n");                    else                        printf("In different gangs.\n");                }            }            else            {                scanf("%d %d",&x, &y);                int xx = _find(x);                int r1 = r;                int yy = _find(y);                int r2 = r;                f[xx] = yy;                if(r1 == r2)                    w[xx] = 1;                else                    w[xx] = 0;            }        }    }    return 0;}