[kuangbin带你飞]专题五 并查集

Wireless Network

裸的并查集，每次修好一个点之后就扫一遍，将它与在它连接范围内的之前就修好的点进行合并，唯一需要注意的点就是可以用点之间距离的平方来避免精度问题

#include 
#include 
#include 

const int MAXN = 10000 + 5;

struct Point {
    int x, y;
} point[MAXN];

struct Union_Find {
	int dad[MAXN];
	int n;

	void init() {
		for (int i = 1; i <= n; i++)
			dad[i] = i;
	}
	Union_Find(int n) {
		this->n = n;
		init();
	}
	int find(int x) {
		if (dad[x] != x) dad[x] = find(dad[x]);
		return dad[x];
	}
	void merge(int x, int y) {
		dad[find(x)] = find(y);
	}
	bool judge(int x, int y) {
		return find(x) == find(y);
	}
};


int n, d;
bool is_fixed[MAXN];

int dis(Point a, Point b) {
    int ans = (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
    return ans;
}

int main()
{
    scanf("%d %d", &n, &d);
    d *= d;     Union_Find uf(n);
    for (int i = 1; i <= n; i++) {
        scanf("%d %d", &point[i].x, &point[i].y);
    }

    char f;
    while (std::cin >> f) {
        if (f == 'O') {
            int x;
            scanf("%d", &x);
            is_fixed[x] = true;
            for (int i = 1; i <= n; i++) {
                if (i == x) continue;
                if (is_fixed[i] && dis(point[i], point[x]) <= d) uf.merge(x, i);
            }
        }
        else {
            int x, y;
            scanf("%d %d", &x, &y);
            uf.judge(x, y) ? puts("SUCCESS") : puts("FAIL");
        }
    }
    return 0;
}

The Suspects

这道题需要询问带元素 $0$ 的集合的元素个数，可以开一个数组 $sum[i]$ ，表示以 $i$ 为根节点的集合的元素个数，因为每个集合的元素个数都保存在根节点上了，所以在合并和输出的时候都要找到根节点来进行操作

#include 
#include 

const int MAXN = 30000 + 2;

int n, m;

struct Union_Find {
	int dad[MAXN], sum[MAXN];
	int n;

	void init() {
		for (int i = 0; i < n; i++) {
			dad[i] = i;
            sum[i] = 1;
		}
	}
	Union_Find(int n) {
		this->n = n;
		init();
	}
	int find(int x) {
		if (dad[x] != x) dad[x] = find(dad[x]);
		return dad[x];
	}
	void merge(int a, int b) {
        int x = find(a);
        int y = find(b);
        if (x != y) {
            dad[y] = x;
            sum[x] += sum[y];
        }
	}
};


int main()
{
    while (scanf("%d %d", &n, &m) != EOF) {
        if (n == 0 && m == 0) break;

        Union_Find uf(n);
        for (int i = 1; i <= m; i++) {
            int k;
            scanf("%d", &k);

            int x, y;
            scanf("%d", &x);
            k--;
            while (k--) {
                scanf("%d", &y);
                uf.merge(x, y);
            }
        }
        printf("%d\n", uf.sum[uf.find(0)]);
    }
    return 0;
}

How Many Tables

每次合并都会少一个桌子，所以维护一个 $sum$ 表示桌子的数量即可

#include 
#include 

const int MAXN = 1000 + 5;

struct Union_Find {
	int dad[MAXN];
	int n, sum;

	void init() {
		for (int i = 1; i <= n; i++)
			dad[i] = i;
        sum = n;
	}
	Union_Find(int n) {
		this->n = n;
		init();
	}
	int find(int x) {
		if (dad[x] != x) dad[x] = find(dad[x]);
		return dad[x];
	}
	void merge(int x, int y) {
		dad[find(x)] = find(y);
        sum--;
	}
	bool judge(int x, int y) {
		return find(x) == find(y);
	}
};

int main()
{
    int t;
    scanf("%d", &t);
    while (t--) {
        int n, m;
        scanf("%d %d", &n, &m);
        Union_Find uf(n);
        for (int i = 1; i <= m; i++) {
            int x, y;
            scanf("%d %d", &x, &y);
            if (!uf.judge(x, y)) uf.merge(x, y);
        }
        printf("%d\n", uf.sum);
    }
    return 0;
}

施工中