并查集

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Words count in article: 647

冗余连接

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class Solution {
public:
    static const int N = 1010;
    int p[N], indegree[N], n;

    int find(int x) { // 查找x的祖宗节点
        if (x != p[x])  p[x] = find(p[x]);
        return p[x];
    }

    bool assertRemove(vector<vector<int>>&edges, int idx) {
        // 建立并查集的过程中检查删除该边后是否还有环
        for (int i = 0; i < n; ++ i) { // 遍历边集
            if (i == idx) continue;  // 排除该边
            if (find(edges[i][0]) == find(edges[i][1]))
                return false;
            p[find(edges[i][0])] = find(edges[i][1]);
        }
        return true;
    }

    vector<int> remove(vector<vector<int>> &edges) {
        
        for (int i = 0; i < n; ++ i) {
            if (find(edges[i][0]) == find(edges[i][1]))
                return edges[i];
            p[find(edges[i][0])] = find(edges[i][1]);
        }
        return {};
    }
    vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {
        n = size(edges);
        vector<int> vec;
        for (int i = 1; i <= n; ++ i) p[i] = i;// 初始化并查集

        // 查看是否有入度为2的点
        for (int i = 0; i < n; ++ i) 
            ++ indegree[edges[i][1]];
        for (int i = n - 1; i >= 0; -- i) {  // 逆序是有多个答案时删除较后的答案
            if (indegree[edges[i][1]]==2)
                vec.push_back(i);//保存节点对
        }

        // 如果有入度为2的点，则只需删除其中一条
        if (vec.size()) {
            if (assertRemove(edges, vec[0])) 
                return edges[vec[0]];
            else 
                return edges[vec[1]];
        }

        // 没有入度为2的点，直接查找成环的边并删除
        return remove(edges);
    }
};

朋友圈

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class Solution {
public:
    int p[210];

    int find(int x) {
        // 递归压缩路径
        // if (x != p[x]) p[x] = find(p[x]);
        // return p[x];       
        // 迭代压缩路径 
        int q = x, tmp;
        while (p[q] != q) q = p[q];  // x的祖宗
        while (x != q) {             // 再从x开始查一遍，顺便把所有经过的节点祖宗更改。
            tmp = p[x], p[x] = q, x = tmp;
        }
        return x;
    }

    int findCircleNum(vector<vector<int>>& M) {
        int n = size(M);
        for (int i = 0; i < n; ++ i) p[i] = i;

        for (int i = 0; i < n; ++ i) {
            for (int j = i; j < n; ++ j) {
                if (M[i][j] == 1) 
                    p[find(i)] = find(j);
            }
        }
        int cnt = 0;
        for (int i = 0; i < n; ++ i){
            if (i == p[find(i)]) ++ cnt;
        } 
        return cnt ;
    }
};

等式方程的可满足性

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class Solution {
public:
    int p[30];

    int find(int x) {
        int q = x, tmp;
        while (q != p[q]) q = p[q];
        while (x != q) tmp = p[x], p[x] = q, x = tmp;
        return x; 
    }

    bool equationsPossible(vector<string>& equations) {
        vector<int> unequ;
        int n = size(equations);
        for (int i = 0; i < 30; ++ i) {
            p[i] = i;
        }

        for (int i = 0; i < n; ++ i) {
            if (equations[i][1] == '=') {
                p[find(equations[i][0]-'a')] = find(equations[i][3]-'a');
            }else {
                unequ.push_back(i);
            }
        }

        for (const auto &i : unequ) {
            if (find(equations[i][0]-'a') == find(equations[i][3]-'a')) 
                return false;
        }
        return true;
    }
};