【面试高频题】难度 3.5/5，综合最短路的 DP 问题

输入：n = 7, roads = [[0,6,7],[0,1,2],[1,2,3],[1,3,3],[6,3,3],[3,5,1],[6,5,1],[2,5,1],[0,4,5],[4,6,2]]

输出：4

解释：从路口 0 出发到路口 6 花费的最少时间是 7 分钟。
四条花费 7 分钟的路径分别为：
- 0 ➝ 6
- 0 ➝ 4 ➝ 6
- 0 ➝ 1 ➝ 2 ➝ 5 ➝ 6
- 0 ➝ 1 ➝ 3 ➝ 5 ➝ 6

输入：n = 2, roads = [[1,0,10]]

输出：1

解释：只有一条从路口 0 到路口 1 的路，花费 10 分钟。

class Solution {
    int N = 210, MOD = (int)1e9+7;
    long INF = (long)1e12;
    int[][] g = new int[N][N];
    int[] in = new int[N];
    long[] dist = new long[N];
    boolean[] vis = new boolean[N];
    int n;
    public int countPaths(int _n, int[][] rs) {
        n = _n;
        for (int[] info : rs) {
            int a = info[0], b = info[1], c = info[2];
            g[a][b] = g[b][a] = c;
        }
        // 朴素 Dijkstra 求解从 0 点到其他点的最短路
        dijkstra();
        // 利用最短路重新建图，并统计入度
        for (int[] info : rs) {
            int a = info[0], b = info[1], c = info[2];
            g[a][b] = g[b][a] = 0;
            if (dist[a] + c == dist[b]) {
                g[a][b] = c; in[b]++;
            } else if (dist[b] + c == dist[a]) {
                g[b][a] = c; in[a]++;
            }
        }
        // 跑拓扑排序统计方案数
        Deque<Integer> d = new ArrayDeque<>();
        for (int i = 0; i < n; i++) {
            if (in[i] == 0) d.addLast(i);
        }
        int[] f = new int[n];
        f[0] = 1;
        while (!d.isEmpty()) {
            int x = d.pollFirst();
            for (int i = 0; i < n; i++) {
                if (g[x][i] == 0) continue;
                f[i] += f[x];
                f[i] %= MOD;
                if (--in[i] == 0) d.addLast(i);
            }
        }
        return f[n - 1];
    }
    void dijkstra() {
        Arrays.fill(dist, INF);
        dist[0] = 0;
        for (int i = 0; i < n; i++) {
            int t = -1;
            for (int j = 0; j < n; j++) {
                if (!vis[j] && (t == -1 || dist[j] < dist[t])) t = j;
            }
            vis[t] = true;
            for (int j = 0; j < n; j++) {
                if (g[t][j] == 0) continue;
                dist[j] = Math.min(dist[j], dist[t] + g[t][j]);
            }
        }
    }
}