{"componentChunkName":"component---src-templates-blog-js","path":"/blog/algorithm/algo_20200320_7/","result":{"data":{"markdownRemark":{"html":"

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Union-Find & Hashing 문제

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노드가 추가될 때마다, 추가된 노드가 포함된 Tree의 크기를 구하라.

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이름(String):index(Integer)로 해싱하여 map에 추가한다.

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Algorithm :

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  1. \n

    두 이름 A, B를 입력받고, map에 포함되지 않은 이름이면(새로운 노드이면),

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    2. \n
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    index로 변환된 A와 B를 합친다. (union)

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    3. \n
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Java Code

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import java.util.HashMap;\nimport java.util.Map;\nimport java.util.Scanner;\n\npublic class Main {\n    static Scanner in = new Scanner(System.in);\n    static int t, f;\n    static Map<String, Integer> map;\n    static int[] groupCnt, parent;\n    static StringBuilder sb = new StringBuilder();\n    public static void main(String[] args) {\n        t = in.nextInt();\n        while(t-- >0) {\n            f = in.nextInt();\n            map = new HashMap<>();\n            groupCnt = new int[f*2];\n            parent = new int[f*2];\n//            노드는 2개씩 주어지므로 전부 다른 노드일 경우 최대 f*2개이다.\n            int idx = 0;\n            while (f-- >0){\n                String a = in.next(); String b = in.next();\n                if(map.get(a)==null) {\n                    parent[idx] = idx;\n                    groupCnt[idx] = 1;\n                    map.put(a, idx++);\n                }\n                if (map.get(b)==null) {\n                    parent[idx] = idx;\n                    groupCnt[idx] = 1;\n                    map.put(b, idx++);\n                }\n                union(map.get(a), map.get(b));\n            }\n        }\n        System.out.println(sb.toString());\n    }\n    public static void union(int a, int b) {\n        int aRoot = find(a), bRoot = find(b);\n        int sum;\n        if(aRoot==bRoot) {\n            sum = groupCnt[aRoot];\n        }\n        else {\n            parent[bRoot] = aRoot;\n            sum = groupCnt[aRoot]+groupCnt[bRoot];\n            groupCnt[aRoot] = sum;\n            groupCnt[bRoot] = sum;\n        }\n        sb.append(sum+\"\\n\");\n\n    }\n    public static int find(int node) {\n        if(node==parent[node]) {\n            return node;\n        }\n        else return parent[node] = find(parent[node]);\n    }\n\n}
","id":"b4dee449-dd4d-5e2f-a1f5-0711238f563c","frontmatter":{"title":"[Algorithm] 백준 4195 : 친구 네트워크 ","date":"2020-03-20T00:00:00.000Z","description":null},"fields":{"slug":"/blog/algorithm/algo_20200320_7/"}}},"pageContext":{"slug":"/blog/algorithm/algo_20200320_7/"}}}