POJ 1703 Find them, Catch them
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题目:
Find them, Catch them
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 31232 Accepted: 9621
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.
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.传送门:点击打开链接
解题思路:
分组并查集(种类并查集).建立两个集合A和B,A中存放和元素a在同一个帮派的元素,B中存放和元素a在不同帮派(一共两个帮派)的元素。对于'D'操作,我们只需把a和b+n放到一个集合,a+n和b放到一个集合即可,因为a和b不在一个帮派,所以a和b+n在一个帮派,同理,a+n和b在一个帮派。对于‘A’操作,我们先判断a和b是否在同一个帮派,即看a和b或者a+n和b+n是否在同一个集合,再看两个元素是否在不同帮派,即看a+n和b或者a和b+n是否在同一个集合,剩下的情况是不确定。
代码:
#include <cstdio>#include <cstring>const int MAXN = 1e5 + 10;int set[MAXN<<1];int find(int p){ if(set[p] < 0) return p; return set[p] = find(set[p]);}void join(int p, int q){ p = find(p), q = find(q); if(p != q) set[p] = q;}int main(){ int t, n, m; scanf("%d", &t); while(t--) { scanf("%d%d", &n, &m); memset(set, -1, sizeof(set)); while(m--) { char s[2]; int a, b; scanf("%s%d%d", s, &a, &b); switch(s[0]) { case 'D': { join(a, b+n); join(b, a+n); break; } case 'A': { if(find(a)==find(b) || find(a+n)==find(b+n)) { printf("In the same gang.\n"); } else if(find(a)==find(b+n) || find(a+n)==find(b)) { printf("In different gangs.\n"); } else { printf("Not sure yet.\n"); } break; } } } }}
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