F - 食物链 POJ - 1182

https://vjudge.net/contest/360957#problem/F

题解

https://blog.csdn.net/niushuai666/article/details/6981689

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 50010
 
struct node
{
    int pre;
    int relation;
};
node p[N];
 
int find(int x) //查找根结点
{
    if(x == p[x].pre)
        return x;
    int temp = p[x].pre; //路径压缩
    p[x].pre = find(temp);
    p[x].relation = (p[x].relation + p[temp].relation) % 3; //关系域更新
    return p[x].pre; //根结点
}
 
int main()
{
    int n, k;
    int ope, a, b;
    int root1, root2;
    int sum = 0; //假话数量
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; ++i) //初始化
    {
        p[i].pre = i;
        p[i].relation = 0;
    }
    for(int i = 1; i <= k; ++i)
    {
        scanf("%d%d%d", &ope, &a, &b);
        if(a > n || b > n) //条件2
        {
            sum++;
            continue;
        }
        if(ope == 2 && a == b) //条件3
        {
            sum++;
            continue;
        }
        root1 = find(a);
        root2 = find(b);
        if(root1 != root2) // 合并
        {
            p[root2].pre = root1;
            p[root2].relation = (3 + (ope - 1) +p[a].relation - p[b].relation) % 3;
        }
        else
        {
            if(ope == 1 && p[a].relation != p[b].relation)
            {
                sum++;
                continue;
            }
            if(ope == 2 && ((3 - p[a].relation + p[b].relation) % 3 != ope - 1))
            {
                sum++;
                continue;}
        }
    }
    printf("%d\n", sum);
    return 0;
}

标签：pre,食物链,int,sum,1182,ope,POJ,relation,root2
来源： https://www.cnblogs.com/SunChuangYu/p/12488964.html