Acwing 240. 食物链

题目描述

思路讲解

代码展示

#include <iostream>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;
}