POJ1703 Find them, Catch them （分类并查集）

Total Submission(s) : 10   Accepted Submission(s) : 3

Problem 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.

 

#include <iostream>#include <algorithm>#include <string.h>#include <stdio.h>#include <map>#include <cmath>#include <queue>using namespace std;const int maxn = 100000;int fa[(maxn<<1) + 10];    //输入a， b 则a， b阵营不同，让 a， b + N 阵营相同，不就解决了这个问题    //由于关系是相对的，所以a + N， b。也要维护这个关系int find(int x) {    if (fa[x] == x)        return x;    fa[x] = find(fa[x]);    return fa[x];}void merge(int x, int y) {    int fx = find(x);    int fy = find(y);    if (fx != fy) {        fa[fy] = fx;    }}void init(int n) {    for (int i = 1; i <= n; ++i) {        fa[i] = i;        fa[i + maxn] = i + maxn;    }}int main() {        //freopen("in.txt", "r", stdin);    int t;    scanf("%d", &t);    while (t--) {        int n, m;        int a, b;        cin >> n >> m;        init(n);        char ch[2];        for (int i = 0; i < m; ++i) {            scanf("%s%d%d", ch, &a, &b);            if (ch[0] == 'A') {                int fa = find(a);                int fb = find(b);                int fbn = find(b + maxn);                if (fa == fbn) {                    printf("In different gangs.\n");                } else if (fa == fb) {                    printf("In the same gang.\n");                } else {                    printf("Not sure yet.\n");                }            } else {                merge(a, b + maxn);                merge(a + maxn, b);            }        }    }}