POJ - 1703 Find them, Catch them（种类并查集）

Find them, Catch them

Time Limit: 1000MS Memory Limit: 10000KB 64bit IO Format: %I64d & %I64u

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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.

题意：输入n和m，代表有n个人和m个询问，n个人分别属于两个帮派，每个询问遵循下面的规则：

A a b要求输出a与b是否在同一个帮派。D a b告诉信息a和b是不同帮派的成员。

这题与种类并查集的经典题目POJ-1182 食物链相似，更简单一些，关系域更新的时候考虑偏移量和向量的加减，可以得出公式。

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int MAXN = 100100;int fa[MAXN];int gang[MAXN];int n, k;void init(){for (int i = 0; i <= n; i++){fa[i] = i;gang[i] = 0;}}int find(int t){if (fa[t] == t)return fa[t];int temp = fa[t];fa[t] = find(temp);gang[t] = (gang[t] + gang[temp]) % 2;return fa[t];}void union_set(int a, int b){int roota = find(a);int rootb = find(b);if (roota != rootb){fa[rootb] = roota;gang[rootb] = (gang[a] + 1 - gang[b]) % 2;}}int solve(int a, int b){int roota = find(a);int rootb = find(b);if (roota != rootb){return 1;}else{if (gang[a] != gang[b])return 2;elsereturn 3;}}int main(){int casen;scanf("%d", &casen);while (casen--){scanf("%d%d", &n, &k);init();char op;int a, b;for (int i = 0; i < k; i++){cin >> op;scanf("%d%d", &a, &b);if (op == 'D'){union_set(a, b);}else{int ret = solve(a, b);if (ret == 1)printf("Not sure yet.\n");else if (ret == 2)printf("In different gangs.\n");elseprintf("In the same gang.\n");}}}}