链接：

题目：

Find them, Catch them

Time Limit: 1000MS

Memory Limit: 10000K

Total Submissions: 22289

Accepted: 6648

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

Sample Output

Not sure yet.In different gangs.In the same gang.

Source

 

本题是POJ 1182的简化版， union_set两个方向合并都可以

代码一次就写对了，但是输出different情况时候少了个s，晕个半死。。。。

 

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;const int maxn=100000+5;int p[maxn], r[maxn];void make_set(){    memset(p, -1, sizeof(p));    memset(r, 0, sizeof(r));}int find_set(int x){    if(p[x]==-1)        return x;    int fx=p[x];    p[x]=find_set(p[x]);    r[x]=(r[x]+r[fx])%2;    return p[x];}void union_set(int x, int y, int d){    int fx=find_set(x), fy=find_set(y);    if(fx==fy) return;#if 0    p[fx]=fy;    r[fx]=(2-r[x]+d+r[y])%2;#else    p[fy]=fx;    r[fy]=(2-r[y]+d+r[x])%2;#endif}int main(){    int T;    int N, M;    scanf("%d", &T);    while(T--) {        scanf("%d%d", &N, &M);        make_set();        char c[20]; int x, y;        for(int i=0;i
         
        
       
      
     

 

另一种写法：

这题是食物链的弱化版，使用相似的思路来做即可——

定义并查集为：

并查集里的元素i-x表示i属于帮派x

同一个并查集的元素同时成立

 

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;const int maxn=100000*2+5;int p[maxn];void make_set(){    memset(p, -1, sizeof(p));}int find_set(int x){    if(p[x]==-1)        return x;    return p[x]=find_set(p[x]);}void union_set(int x, int y){    int fx=find_set(x), fy=find_set(y);    if(fx==fy) return;    p[fx]=fy;}bool same(int x, int y){    return find_set(x)==find_set(y);}int main(){    int T;    int N, M;    scanf("%d", &T);    while(T--) {        scanf("%d%d", &N, &M);        make_set();        char c[20]; int x, y;        for(int i=0;i
         
        
       
      
     

稍微扯远一点，最近看完了《数学之美》这本书，里面提到了最大熵模型，也就是“包含所有可能性”的模型。这道题目其实归根结底就是保留了所有可能性，我们只知道x和y不属于同一集合，但我们不能确定究竟x属于集合A还是集合B，于是我们保留所有可能性，对x-A和x-B都做了一次记录。

上面这句话转自：