Acwing 240. 食物链

题目描述

思路讲解

代码展示

#include 

using namespace std;

const int N = 50010;

int n, m;
int p[N], d[N]; //p[]是并查集的father,d[]是距离

int find(int x) {
    if (p[x] != x) {  //如果说x不是树根的话
        int t = find(p[x]);
        d[x] += d[p[x]];
        p[x] = t;
    }
    return p[x];
}

int main() {
    scanf("%d%d", &n, &m);

    for (int i = 1; i <= n; i++) p[i] = i;  //初始化

    int res = 0;
    while (m--) {
        int t, x, y;
        scanf("%d%d%d", &t, &x, &y);

        if (x > n || y > n) res++;
        else {
            int px = find(x), py = find(y);
            if (t == 1) {
                if (px == py && (d[x] - d[y]) % 3) res++;
                else if (px != py) {
                    p[px] = py;
                    d[px] = d[y] - d[x];
                }
            } else {
                if (px == py && (d[x] - d[y] - 1) % 3) res++;
                else if (px != py) {
                    p[px] = py;
                    d[px] = d[y] + 1 - d[x];
                }
            }
        }
    }

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

    return 0;
}