UVa 10158 War

Problem B: War

    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

一道并查集问题，于POJ1182属于同一类型，但更简单一些。

用relation[x]表示在并查集中，x结点与根结点的关系，relation[x]=0表示x与根节点是朋友关系，relation[x]=1表示x与根节点是敌人关系，并查集查找和合并操作时的关系转换关系

#include<iostream>
#include<cstdio>
#include<cstring>
#define MAX_N 10050

using namespace std;

int par[MAX_N],relation[MAX_N];

void init(int n)
{
    memset(relation,0,sizeof(relation));
    for(int i=0;i<n;i++)
        par[i]=i;
}

int Find(int x)
{
    if(par[x]==x)
        return x;
    int pre_father=par[x];
    par[x]=Find(par[x]);
    relation[x]=(relation[x]+relation[pre_father]+2)%2;
    return par[x];
}

void Unite(int a,int b,int x,int y,int r)
{
    par[a]=b;
    relation[a]=(2+relation[y]+relation[x]+r)%2;
}

int main()
{
    int n;

    while(scanf("%d",&n)==1)
    {
        int c,x,y;

        init(n);

        while(scanf("%d %d %d",&c,&x,&y)==3&&(c||x||y))
        {
            if(c==1)
            {
                if(Find(x)!=Find(y))
                    Unite(Find(x),Find(y),x,y,0);
                else if((relation[x]-relation[y]+2)%2!=0)
                    puts("-1");
            }
            else if(c==2)
            {
                if(Find(x)!=Find(y))
                    Unite(Find(x),Find(y),x,y,1);
                else if((relation[x]-relation[y]+2)%2!=1)
                    puts("-1");
            }
            else if(c==3)
            {
                if(Find(x)==Find(y)&&(relation[x]-relation[y]+2)%2==0)
                    puts("1");
                else
                    puts("0");
            }
            else if(c==4)
            {
                if(Find(x)==Find(y)&&(relation[x]-relation[y]+2)%2==1)
                    puts("1");
                else
                    puts("0");
            }
        }
    }

    return 0;
}