poj 1703 并查集解决分组问题

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.

题意： 一共有2个帮派，D a b表示a和b属于两个不同的帮派，A a b表示询问a和b是否属于一个帮派，或者无法确定。

在并查集树中，权值只有0和1，0表示此结点和父节点处于同一帮派，1则相反。

#include<iostream>#include<cmath>#include<cstring>#include<cstdio>#include<vector>#include<map>#include<queue>#include<algorithm>#include<sstream>#define inf 0x3f3f3f3fusing namespace std;int pre[100010];int flag[100010];int findset(int x){    if(pre[x]!=x)    {        int t=findset(pre[x]);        flag[x]=flag[x]^flag[pre[x]];        //flag[pre[x]]表示x的父节点和根节点的关系，，flag[x]表示x结点和父节点关系，        //如果flag[x]和flag[pre[x]]一个为0一个为1，表明x和根节点一定不同为1        //如果flag[x]和flag[pre[x]]都为0，表明x和父节点、根节点都相同        //如果flag[x]和flag[pre[x]]都为1，表明x和父节点不同，父节点和根节点不同，x和根节点相同        return pre[x]=t;    }    return x;}void union_set(int u,int v){    int x=findset(u);    int y=findset(v);    if(x!=y)    {        pre[x]=y;        flag[x]=~(flag[u]^flag[v]);        //flag[u]为u与根节点x关系，flag[v]为v与根节点y关系        //得到x与y的关系    }}int main(){    int T;    cin>>T;    while(T--)    {        int n,q;        cin>>n>>q;        for(int i=1;i<=n;i++)            pre[i]=i;        memset(flag,0,sizeof(flag));//初始权值都为0，这样初次加入的时候第32行操作子节点的权值为1        while(q--)        {            char c;            cin>>c;            int u,v;            scanf("%d%d",&u,&v);            if(c=='D')            {                union_set(u,v);            }            else            {                int x=findset(u);                int y=findset(v);                if(x==y)                {                    if(flag[u]==flag[v])                        cout<<"In the same gang."<<endl;                    else                        cout<<"In different gangs."<<endl;                }                else                {                    cout<<"Not sure yet."<<endl;                }            }        }    }    return 0;}