poj1703_Find them, Catch them_并查集

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 42451		Accepted: 13059

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.

Source

POJ Monthly--2004.07.18

 1 #include <iostream>
 2 #include <cstdio>
 3 
 4 #define MAX_N 1000000+5
 5 
 6 using namespace std;
 7 
 8 int par[MAX_N];//父节点
 9 int depth[MAX_N];//深度
10 
11 void init(int n){
12    for(int i=0;i<=n;i++){
13        par[i]=i;
14        depth[i]=1;
15    }
16 }
17 int find_father(int t){
18    if(t==par[t]){
19        return t;
20    }else{
21        return par[t]=find_father(par[t]);
22        //实现了路径压缩
23    }
24 }
25 void unite(int t1,int t2){
26    int f1=find_father(t1);
27    int f2=find_father(t2);
28    if(f1==f2){
29        return ;
30    }
31    if(depth[f1]<depth[f2]){
32        par[f1]=f2;
33    }else{
34        par[f2]=f1;
35        if(depth[f1]==depth[f2]){
36            depth[f1]++;
37            //记录深度
38        }
39    }
40 }
41 
42 bool same(int x,int y){
43     return find_father(x)==find_father(y);
44 }
45 
46 
47 int main()
48 {
49     int t;
50     int n,m;
51     char c;
52     int a,b;
53     scanf("%d",&t);
54     while(t--){
55         scanf("%d %d",&n,&m);
56         init(n*2);
57         while(m--){
58             getchar();
59             scanf("%c %d %d",&c,&a,&b);
60             if(c=='D'){
61                 unite(a,b+n);
62                 unite(a+n,b);
63             }else{
64                 if(same(a,b+n)){
65                     printf("In different gangs.\n");
66                     continue;
67                 }
68                 if(same(a,b)){
69                    printf("In the same gang.\n");
70                    continue;
71                 }else{
72                    printf("Not sure yet.\n");
73                    continue;
74                 }
75             }
76         }
77     }
78     return 0;
79 }
80 
81

 

转载于:https://www.cnblogs.com/TWS-YIFEI/p/6039313.html