POJ 1182 食物链

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题意就不讲解了，很经典的一道题目，这是链接：<span style="font-family: Arial, Helvetica, sans-serif;">http://poj.org/problem?id=1182</span>

题解：利用并查集，注意考虑合并时的关系更新

</pre><pre name="code" class="cpp">#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<algorithm>#include <iostream>#include <queue>#include <stack>#include <vector>#include <map>#include <set>#define INF 0x3f3f3f3f#define maxn 50010using namespace std;int parent[50010], relation[50010]; //parent[x]是x的父节点 relation[x]是x与根节点的关系int N, K, X, Y, D, ans = 0;void init(){    for(int i = 0; i < N; i++)    {        parent[i] = i;        relation[i] = 0;    }}                              //初始化数组int find(int x){    if(x != parent[x])    {        int temp = parent[x];        parent[x] = find(temp);                     //必须先从父节点更新        relation[x] = (relation[x] + relation[temp]) % 3; //不断更新当前节点与根节点关系    }    return parent[x];}                          //寻找x的根节点void Union(){    int px = find(X);    int py = find(Y);    parent[py] = px;    relation[py] = (D - 1 + 3 - relation[Y] + relation[X]); //更新Y根节点与总的根节点关系}                                              //合并X,Yint main(){    scanf("%d%d", &N, &K);    init();    while(K--)    {        scanf("%d%d%d", &D, &X, &Y);        if(X > N || Y > N)            ans++;        else if(D == 2 && X == Y)            ans++;        else        {            if(find(X) == find(Y))            {                if((3 - relation[X] + relation[Y]) % 3 != D - 1) //3-部分是X与根节点关系   relation[Y]是根节点与X关系  加起来就是X与Y关系                    ans++;            }            else                Union();        }    }    printf("%d\n",ans);    return 0;}

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