ybtoj并查集例题4食物链

Posted 2021-05-16 SSL_ZZL

【例题4】食物链

• • • Link
    • 解题思路
    • Code

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Link

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题目

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解题思路

设x+n是吃x的，x吃x+2*n，y同上
对于给出的两种关系，我们把一种可能的a、b、c列出来

PS：此处的a、b、c只是代表它们的关系，代表了谁吃谁、谁和谁同类，并不是一定是这种排列

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Code

#include <iostream>
#include <cstdio>

using namespace std;

int fa[150100], n, m, c, x, y, ans;
int x0, y0, x1, y1, x2, y2, flag;

int find(int x) {
	if (fa[x] == x) return x;
	fa[x] = find(fa[x]);
	return fa[x];
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= 3 * n; i++)
		fa[i] = i;
	for (int i = 1; i <= m; i++) {
		scanf("%d %d %d", &c, &x, &y);
		if (x > n || y > n || (c == 2 && x == y)) {  //先把基础的假话做了
			ans++;
			continue;
		}
		
		flag = 1;
		x0 = find(x), y0 = find(y);                   //x(x0), y(y0)
		x1 = find(x + n), y1 = find(y + n);           //x+n(x1), y+n(y1)             x/y + s*n (x/y s)这样还挺对应的
		x2 = find(x + 2 * n), y2 = find(y + 2 * n);   //x+2*n(x2), y+2*n(y2)
		
		if (c == 1) {  //保险起见，我把所有的对应
			if (x1 == y0 || x1 == y2 || x0 == y1 || x0 == y2 || x2 == y1 || x2 == y0) {
				ans++, flag = 0;
			}
		} else {
			if (x1 == y1 || x1 == y0 || x0 == y0 || x0 == y2 || x2 == y1 || x2 == y2) {
				ans++, flag = 0;
			}
		}
		
		if (!flag) continue;
		
		if (c == 1) {   //把同类并查集
			fa[x0] = y0, fa[x1] = y1, fa[x2] = y2;
		} else {
			fa[x0] = y1, fa[x2] = y0, fa[x1] = y2;
		}
	}
	printf("%d", ans);
}