poj1703 Find them, Catch them（并查集的应用）

Find them, Catch them

 

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 32225		Accepted: 9947

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.
 
题目大意：N 表示共有 N 个帮派；下面有 M 条个询问
        有两个帮派，现在告诉你一些互相敌对的点对，然后让你判断某两个点之间的关系，并首先判两个点关系是否确定；
        用并查集的思想只要两个点在一个集合里就可以了，直接用并查集然后时候在同一个集合；
AC代码：

 #include<iostream>
 #include<algorithm>
 #include<cstdio>
 using namespace std;
 // 0 代表木有关系，1表示不同类，2代表同类
 const int N = 1e5+;
 int m;
 int f[N],c[N];
 int init()
 {
     for(int i=; i<=m; i++)
     {
         f[i] = i;
         c[i] = ;
     }
 }
 int xfind(int x)
 {
     if(f[x] == x)
        return x;
        int t = f[x];
        f[x] =xfind(f[x]);
        c[x] = (c[x]+c[t])%;
        return f[x];
 }
 int solve(char ch,int x,int y )
 {
     int fx = xfind(x);
     int fy = xfind(y);
     if(fx != fy)
     {
         if(ch == 'A') return ; //返回值0，表示不确定
         f[fx] =fy;
         if((c[x]+c[y])% ==)
             c[fx] = ;
     }
     else
     {
         if(ch == 'A')
            {
                if(c[x] != c[y]) return ;    //返回值1，表示在不同的帮派
                 if(c[x] == c[y]) return ;  //返回值2，表示在相同的帮派
            }
     }
     if(ch == 'D') return ;//这种情况不同在考虑的
 }
 int main( )
 {
     int T,n,x,y;
     char s;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d %d",&m,&n);
         init();         //初始化并查集
         for(int i=; i<n; i++)
         {
             scanf(" %c %d %d",&s,&x,&y);
             int flag  = solve(s,x,y);
             if(flag == ) printf("Not sure yet.\n");
             else if(flag == ) printf("In different gangs.\n");
             else if(flag == ) printf("In the same gang.\n");
         }
     }
     return ;
 }