【poj1182】【经典】【带权并查集】【关系判断矛盾】

【链接】

给出几条关系,判断有几个是矛盾的

i-x代表“i属于x类”

并查集内同组所有元素代表的情况同时发生或者同时不发生

分类:

同类:合并x-A和y-A,x-B和y-B,x-C和y-C

捕食关系(x吃y):合并x-A和y-B,x-B和y-C,x-C和y-A

【代码】

#include<cstdio>
using namespace std;
const int maxn = 150010;
int fa[maxn];

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

void mix(int x, int y) {
	int fx = find(x);
	int fy = find(y);
	if (fx != fy)fa[fx] = fy;
}

int main() {
	int n, k;
	scanf("%d%d", &n, &k);
		for (int i = 1; i <3* n; i++) {
			fa[i] = i;
		}
		int ans = 0;
		while (k--) {
			int d, x, y;
			scanf("%d%d%d", &d, &x, &y);
			if (x<=0 || y<=0 || x > n || y > n) {
				ans++;
				continue;
			}
			if (d == 1) {
				if (find(x) == find(y+n)||find(x)==find(y+2*n)) {
					ans++;
				}
				else {
					mix(x, y);
					mix(x + n, y + n);
					mix(x + 2 * n, y + 2 * n);
				}
			}
			else {
				if (find(x)==find(y+2*n)||find(x)==find(y)) {
					ans++;
				}
				else {
					mix(x, y + n);
					mix(x + n, y + 2 * n);
					mix(x + 2 * n, y);
				}
			}
		}
		printf("%d\n", ans);
}

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转载自blog.csdn.net/running_acmer/article/details/82846161