Find them, Catch them POJ

Find them, Catch them

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 46665 Accepted: 14367

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.

Source

POJ Monthly--2004.07.18

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Link: http://poj.org/problem?id=1703

题意：有两个团伙，给出两种操作：

    A X Y：询问 X Y 是否为同伙 ;

            D X Y：明确指出X Y为不同伙 ;

思路：典型的并查集思想，但指令给出的是不同伙（根节点不同），将两个团伙分成两个集，但接收到D指令时仍              然使用Combine（）合并，使用+n区别开 ;

AC Code：

#include<cstdio>using namespace std;const int maxn=500000+50;int fa[maxn],rank[maxn];void init(int n){    for(int i=1;i<=n;i++){        fa[i]=i;        rank[i]=0;    }}int find(int x){    return fa[x]==x?x:fa[x]=find(fa[x]);}void myunion(int a,int b){    a=find(a); b=find(b);    if(a==b) return ;    if(rank[a]>rank[b]) fa[b]=a;    else{        fa[a]=b;        if(rank[a]==rank[b]) ++rank[b];    }}bool issame(int a,int b){    return find(a)==find(b);}int main(){    int t;    scanf("%d",&t);    while(t--){        int n,m;        scanf("%d%d",&n,&m);        init(n*2);        while(m--){            char str[5];            int a,b;            scanf("%s%d%d",str,&a,&b);            //printf("%s %d %d\n",str,a,b);            if(str[0]=='D'){                myunion(a,b+n);                myunion(a+n,b);            }            else{                if(issame(a,b)) printf("In the same gang.\n");                else if(issame(a+n,b)) printf("In different gangs.\n");                else printf("Not sure yet.\n");            }        }    }    return 0;}