Find them, Catch them poj 1703

最新推荐文章于 2022-07-25 21:18:41 发布

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最新推荐文章于 2022-07-25 21:18:41 发布

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分类专栏：并查集

本文链接：https://blog.youkuaiyun.com/qq_34952846/article/details/53423811

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本文介绍了一种通过并查集算法解决两个帮派成员归属问题的方法。针对给出的两个帮派成员编号，算法能够判断他们是否属于同一帮派，或者由于信息不足暂时无法判断。该算法适用于处理大量成员及消息的情况。

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Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 42775		Accepted: 13164

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

Sample Output

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

有两个帮派，有两种操作 D a b表示a 和 b不是一个帮派；A a b 表示询问a b是否是一个帮派，若至此还不确定，输出“Not sure yet”。

代码：

#include <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;

int b[100005];
int a[100005];

int find(int x)
{
    if ( b[x] == x )
        return x;
    int temp = b[x];
    b[x] = find(b[x]);
    a[x] = a[x]^a[temp];
    return b[x];
}

void group(int x, int y)
{
    int fx = find(x);
    int fy = find(y);
    b[fx] = fy;
    a[fx] = !(a[x]^a[y]);
}

int main()
{
    int n, m, i;
    int T;
    char ch[5];
    int x, y;
    scanf ( "%d", &T );
    while ( T-- )
    {
        memset ( a, 0, sizeof(a) );
        scanf ( "%d %d", &n, &m );
        for ( i = 1;i <= n; i++ )
            b[i] = i;
        for ( i = 0;i < m; i++ )
        {
            scanf ( "%s %d %d", ch, &x, &y );
            if ( ch[0] == 'A' )
            {
                if ( find(x) != find(y) )
                {
                    printf ( "Not sure yet.\n" );
                }
                else
                {
                    if ( a[x] == a[y] )
                        printf ( "In the same gang.\n" );
                    else
                        printf ( "In different gangs.\n" );
                }
            }
            else
            {
                group(x, y);
            }
        }
    }
}

代码菜鸟，如有错误，请多包涵！！！

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