UVA 10158 War（并查集）
A war is being lead between two countries, A and B. As a loyal citizen of C, you decide to help your country’s espionage by attending the peace-talks taking place these days (incognito, of course). There are n people at the talks (not including you), but you do not know which person belongs to which country. You can see people talking to each other, and through observing their behaviour during their occasional one-to-one conversations, you can guess if they are friends or enemies. In fact what your country would need to know is whether certain pairs of people are from the same country, or they are enemies. You may receive such questions from C’s government even during the peace-talks, and you have to give replies on the basis of your observations so far. Fortunately nobody talks to you, as nobody pays attention to your humble appearance.
Abstract
Now, more formally, consider a black box with the following operations:
setFriends(x, y) shows that x and y are from the same country
setEnemies(x, y) shows that x and y are from different countries
areFriends(x, y) returns true if you are sure that x and y are friends
areEnemies(x, y) returns true if you are sure that x and y are enemies
The first two operations should signal an error if they contradict with your former knowledge. The two relations ‘friends’ (denoted by ~) and ‘enemies’ (denoted by *) have the following properties:
~ is an equivalence relation, i.e.
1. If x ~ y and y ~ z then x ~ z (The friends of my friends are my friends as well.)
2. If x ~ y then y ~ x (Friendship is mutual.)
3. x ~ x (Everyone is a friend of himself.)
* is symmetric and irreflexive
4. If x * y then y * x (Hatred is mutual.)
5. Not x * x (Nobody is an enemy of himself.)
Also
6. If x * y and y * z then x ~ z (A common enemy makes two people friends.)
7. If x ~ y and y * z then x * z (An enemy of a friend is an enemy.)
Operations setFriends(x, y) and setEnemies(x, y) must preserve these properties.
Input
The first line contains a single integer, n, the number of people.
Each of the following lines contains a triple of integers, c x y, where c is the code of the operation:
c = 1, setFriends
c = 2, setEnemies
c = 3, areFriends
c = 4, areEnemies
and x and y are its parameters, which are integers in the range [0, n), identifying two (different) people. The last line contains 0 0 0.
All integers in the input file are separated by at least one space or line break.
Output
For every ‘areFriends’ and ‘areEnemies’ operation write 0 (meaning no) or 1 (meaning yes) to the output. Also for every ‘setFriends’ or ‘setEnemies’ operation which contradicts with previous knowledge, output a –1 to the output ; note that such an operation should produce no other effect and execution should continue. A successful ‘setFriends’ or ‘setEnemies’ gives no output.
All integers in the output file must be separated by at least one space or line break.
Constraints
n < 10000, the number of operations is unconstrained.
Sample Input
10
1 0 1
1 1 2
2 0 5
3 0 2
3 8 9
4 1 5
4 1 2
4 8 9
1 8 9
1 5 2
3 5 2
0 0 0
Sample Output
1
0
1
0
0
-1
0
Input
Output
Sample Input
Sample Output
Hint
这个是稍微有点难的并查集。因为多了敌对关系。所以开辟了2n的数组来存放。0-n代表结盟集合,n-2n代表敌对集合.
#include <iostream>
#define maxn 10005*2
using namespace std;
int f[maxn];
int find(int x)
{
if(x!=f[x])
f[x]=find(f[x]);
return f[x];
}
int init(int n)
{
int i;
//0-n代表结盟集合,n-2n代表敌对集合.
for(i=1;i<=n;i++)
{
f[i]=i;
f[i+n]=i+n;
}
}
int main()
{
int n;
cin>>n;
init(n);//初始化
int s,a,b;
while(cin>>s>>a>>b&&s||a||b)
{
int a1=find(a);
int a2=find(a+n);
int b1=find(b);
int b2=find(b+n);
if(s==1)//表示a和b来自同一个国家
{
if(a1==b2)//如果前提条件已证明a和b来自不同国家
cout<<"-1"<<endl;
else
{
f[a1]=b1;
f[a2]=b2;
}
}
if(s==2)//表示a和b来自不同的国家
{
if(a1==b1)//如果前提条件已证明a和b来自同一个国家
cout<<"-1"<<endl;
else
{
f[a1]=b2;
f[a2]=b1;
}
}
if(s==3)//判断a和b是否是朋友
{
if(a1==b1)
cout<<"1"<<endl;
else
cout<<"0"<<endl;
}
if(s==4)//判断a和b是否是敌人
{
if(a1==b2)
cout<<"1"<<endl;
else
cout<<"0"<<endl;
}
}
}