Poj 1703 Find them, Catch them （并查集）

Find them, Catch them

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions:48184 Accepted: 14849

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."

主要是存储那几个数在相反的集合中。

合并的时候合并两次将两个数分别与另一个数的相反数合并到一起。

查询的时候查询 两个数 以及 其中一个数的相反数是否在一个父亲节点下。

#include<stdio.h>#include<string.h>using namespace std;#define N 100005int Opp[N];int pre[N];int Find(int x){    if(pre[x]!=x)    {        return pre[x]=Find(pre[x]);    }    return x;}void Merge(int x,int y){    int X=Find(x);    int Y=Find(y);    if(X!=Y)    {        pre[X]=Y;    }    return;}void Init(){    for(int i=0; i<N; i++)    {        pre[i]=i;        Opp[i]=0;    }}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n,m;        Init();        scanf("%d%d",&n,&m);        while(m--)        {            char op[3];            int x,y;            scanf("%s %d %d",op,&x,&y);            if(op[0]=='D')            {                if(Opp[x]==0)                    Opp[x]=y;                if(Opp[y]==0)                    Opp[y]=x;                Merge(x,Opp[y]);                Merge(y,Opp[x]);            }            else            {                int X=Find(x),Y=Find(y);                int Opp_Y=Find(Opp[y]);                if(X == Y)                {                    printf("In the same gang.\n");                }                else if(X == Opp_Y)                {                    printf("In different gangs.\n");                }                else printf("Not sure yet.\n");            }        }    }}