POJ-1703 Find them, Catch them 经典并查集

题目链接

题目大意：在这个城市里有两个黑帮团伙，现在给出N个人，问任意两个人他们是否在同一个团伙

输入D x y代表x于y不在一个团伙里

输入A x y要输出x与y是否在同一团伙或者不确定他们在同一个团伙里

与poj1182 2492类似

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<math.h>
#include<functional>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int maxn = 100005;
const int inf = 1<<29;
int n,m;
int p[maxn],g[maxn];  //g 0表示相同 1表示不同

int find( int x )
{
if( p[x] == x )	return x;
int tmp = p[x];
p[x] = find(p[x]);
g[x] = (g[x]+g[tmp])%2;
return p[x];
}
void merge( int x,int y )
{
int fx = find(x);
int fy = find(y);
p[fy] = fx;
g[fy] = ( g[y] + g[x] + 1 )%2;
}
void init()
{
for( int i = 0; i <= n; i ++ ){
p[i] = i;
g[i] = 0;
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("data.txt","r",stdin);
#endif
char ch;
int cas,x,y;
scanf("%d",&cas);
while( cas --  ){
scanf("%d%d ",&n,&m);
init();
for( int i = 0; i < m; i ++ ){
scanf("%c%d%d ",&ch,&x,&y);
if( ch == 'A' ){
if( find(x) == find(y) ){
if( g[x] == g[y] )
puts("In the same gang.");
else puts("In different gangs.");
}
else puts("Not sure yet.");
}
else if( ch == 'D' ){
merge( x,y );
}
}
}
return 0;
}