poj1703 并查集

Find them, Catch them

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 41067 Accepted: 12631

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.

题目就是说给定M句话，N个人，请你判断其中的某两位是不是一个帮派的人

这里运用的是并查集，用两个集合来表示的，因此集合比之前的扩大了一倍

其中扩大的一倍表示它的影子，如  a  +  N，就表示为 a  的影子

然后判断：要是 b 和 a 的影子是一个帮派的人，那么a  b不是一个帮派的

判断a  b是一个帮派的人那肯定就不用说了啊   其余的自然就是不能判断的咯

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int maxx=100005;int father[maxx<<1];int find_it(int x){    int tempx=x,t;//-------------找根节点-------------------//    while(father[tempx])        tempx=father[tempx];//-------------路径压缩-------------------//    while(tempx!=x)    {        t=father[x];//记录父亲        father[x]=tempx;//压缩        x=t;//回溯    }    return tempx;}//----------------合并-------------------//void unite(int x,int y){    x=find_it(x);    y=find_it(y);    if(x!=y)        father[x]=y;}//---------------------------------------//bool same(int x,int y){    return find_it(x)==find_it(y);}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int N,M;        scanf("%d%d",&N,&M);        memset(father,0,sizeof(father[0])*(2*N+1));        while(M--)        {            char ch[5];            int a,b;            scanf("%s%d%d",&ch,&a,&b);            if(ch[0]=='D'){                unite(a,b+N);                unite(a+N,b);            }            else{                if(same(a,b+N))                    printf("In different gangs.\n");                else if(same(a,b))                    printf("In the same gang.\n");                else                    printf("Not sure yet.\n");            }        }    }    return 0;}