POJ 1182 食物链（种类并查集）

题目：http://poj.org/problem?id=1182

 

#include <iostream>

using namespace std;



const int M = 50000 + 10;



struct Node

{

    int father;

    int kind;

}node[M];



void Init(int n)

{

    for (int i = 1; i <= n; i++)

    {

        node[i].father = i;

        node[i].kind = 0;

    }



}

//

int Find(int x)

{

    if (x == node[x].father)

    {

        return x;

    }



    int temp = node[x].father;//temp保存原来的老爸



    node[x].father = Find(node[x].father);



    node[x].kind = (node[x].kind + node[temp].kind) % 3;//////////////



    return node[x].father;

}



//

void Union(int rootx, int rooty, int x, int y, int k)//k == 0:a==b  k == 1:a > b

{



    node[rooty].father = rootx;

    node[rooty].kind = (-k +(node[x].kind - node[y].kind)+3)%3;///////////////



}

//(node[a].kind - node[b].kind + 3) % 3 == 1表示a > b

//node[a].kind == node[b].kond 表示 a == b

//为了方便操作根结点的种类是0

//与根结点直接相连的节点与根结点的关系是正确的

//在同一颗树上的结点表示存在关系



int main()

{

    int n, m;

    scanf("%d%d", &n, &m);//这题只有一组数据



    int ans = 0;



    Init(n);



    while (m--)

    {

        int kind, a, b;

        scanf("%d%d%d", &kind, &a, &b);

        if (a > n || b > n)

        {

            ans++;

            continue;

        }



        if (kind == 2 && a == b)

        {

            ans++;

            continue;

        }



        int roota = Find(a);

        int rootb = Find(b);

        if (roota != rootb)

        {

            Union(roota, rootb, a, b, kind - 1);

        }

        else

        {

            if (kind == 1 && node[a].kind != node[b].kind)

            {

                ans++;

            }



            if (kind == 2 && (node[a].kind - node[b].kind + 3) % 3 != 1)

            {

                ans++;

            }

        }



    }

    printf("%d\n", ans);





    return 0;

}