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백준 1238 파티

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문제 정의 : Directed Graph에서, 모든 노드에 대해 X에 방문했다가 다시 돌아오는 경로의 최소 비용을 구하라.

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(1) 모든 노드에서 X로 가는 최소 비용 +\n(2) X에서 모든 노드로 가는 최소 비용의 합.

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Java Code

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import java.util.LinkedList;\nimport java.util.List;\nimport java.util.PriorityQueue;\nimport java.util.Scanner;\n\nclass edge_Party {\n    int adj, t;\n    edge_Party(int adj, int t) {\n        this.adj = adj; this.t = t;\n    }\n}\nclass node_Party implements Comparable<node_Party> {\n    int v, t;\n    node_Party(int v, int t) {\n        this.v = v; this.t = t;\n    }\n\n    @Override\n    public int compareTo(node_Party o) {\n        if(o.t<this.t) return 1;\n        else if(o.t>this.t) return -1;\n        else return 0;\n    }\n}\n\npublic class Main {\n    static Scanner in = new Scanner(System.in);\n    static int n, m, x;\n    static int[][] d = new int[1001][2];\n    static int[] visited = new int[1001];\n    static PriorityQueue<node_Party> pq = new PriorityQueue<>();\n    static PriorityQueue<node_Party> pq_reverse = new PriorityQueue<>();\n    static List<edge_Party>[] edges = new List[1001];\n    static List<edge_Party>[] edges_reverse = new List[1001];\n    public static void main(String[] args) {\n        n = in.nextInt(); m = in.nextInt(); x = in.nextInt();\n        for(int i=1;i<=n;i++) {\n            edges[i] = new LinkedList<>();\n            edges_reverse[i] = new LinkedList<>();\n            d[i][0] = Integer.MAX_VALUE;\n            d[i][1] = Integer.MAX_VALUE;\n        }\n        for(int i=0;i<m;i++) {\n            int v1 = in.nextInt(); int v2 = in.nextInt(); int t = in.nextInt();\n            edges[v1].add(new edge_Party(v2, t));\n            edges_reverse[v2].add(new edge_Party(v1, t));\n            if(v1==x) {\n                d[v2][0] = t;\n                pq.add(new node_Party(v2, t));\n            }\n            else if(v2==x) {\n                d[v1][1] = t;\n                pq_reverse.add(new node_Party(v1, t));\n            }\n        }\n//        1. x에서 모든 정점으로 가는 최소 비용 (d[node][0], edges)\n//        2. 모든 정점에서 x로 가는 최소 비용 (d[node][1], edges_reverse)\n        int cnt = 1;\n        while (cnt<n) {\n            node_Party curr = pq.poll();\n            if(visited[curr.v]==0) {\n                for(int i=0;i<edges[curr.v].size();i++) {\n                    int adj = edges[curr.v].get(i).adj;\n                    int time = edges[curr.v].get(i).t;\n\n                    if(adj!=x) {\n                        d[adj][0] = Math.min(d[adj][0], d[curr.v][0]+time);\n                        pq.add(new node_Party(adj, d[adj][0]));\n                    }\n                }\n                visited[curr.v] = 1;\n                cnt++;\n            }\n\n        }\n        for(int i=1;i<=n;i++) {\n            visited[i] = 0;\n        }\n        cnt = 1;\n        while (cnt<n) {\n            node_Party curr = pq_reverse.poll();\n            if(visited[curr.v]==0) {\n                for(int i=0;i<edges_reverse[curr.v].size();i++) {\n                    int adj = edges_reverse[curr.v].get(i).adj;\n                    int time = edges_reverse[curr.v].get(i).t;\n                    if(adj!=x) {\n                        d[adj][1] = Math.min(d[adj][1], d[curr.v][1]+time);\n                        pq_reverse.add(new node_Party(adj, d[adj][1]));\n\n                    }\n                }\n                visited[curr.v] = 1;\n                cnt++;\n            }\n        } \n        int res = 0;\n        for(int i=1;i<=n;i++) {\n            if(i!=x) {\n                res = Math.max(res, d[i][0]+d[i][1]);\n            }\n        }\n        System.out.println(res);\n    }\n}
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