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882. 细分图中的可到达节点

时间:2022-11-26 16:14:45浏览次数:61  
标签:882 dist idx int tt hh 图中 new 节点

882. 细分图中的可到达节点

class Solution {
     int N = 3010, M = 20010, INF = 0x3f3f3f3f;
    int[] h = new int[N];
    int[] e = new int[M];
    int[] w = new int[M];
    int[] ne = new int[M];
    int idx;
    int[] dist = new int[N];
    int[] q = new int[N];
    boolean[] st = new boolean[N];
    

    void add(int a, int b, int c) {
        e[idx] = b;
        w[idx] = c;
        ne[idx] = h[a];
        h[a] = idx++;
    }

    void spfa() {
        Arrays.fill(dist, INF);
        dist[0] = 0;
        int hh = 0, tt = 1;
        q[0] = 0;
        while (hh != tt) {
            int t = q[hh++];
            if (hh == N) hh = 0;
            st[t] = false;
            for (int i = h[t]; ~i != 0; i = ne[i]) {
                int j = e[i];
                if (dist[j] > dist[t] + w[i]) {
                    dist[j] = dist[t] + w[i];
                    if (!st[j]) {
                        q[tt++] = j;
                        if (tt == N) tt = 0;
                        st[j] = true;
                    }
                }
            }
        }
    }

    public int reachableNodes(int[][] edges, int maxMoves, int n) {
        Arrays.fill(h, -1);
        idx = 0;
        for (int[] edge : edges) {
            int a = edge[0], b = edge[1], c = edge[2];
            add(a, b, c + 1);
            add(b, a, c + 1);
        }
        spfa();
        int res = 0;
        for (int i = 0; i < n; i++) {
            if (dist[i] <=maxMoves) {
                res ++;
            }
        }
        for (int[] edge : edges) {
            int a = edge[0], b = edge[1], c = edge[2];
            int x = Math.max(0, maxMoves - dist[a]), y = Math.max(0, maxMoves -dist[b]);
            res += Math.min(x + y, c);
        }
        return res;
    }
}

标签:882,dist,idx,int,tt,hh,图中,new,节点
From: https://www.cnblogs.com/eiffelzero/p/16927595.html

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