题意

给定一个包含 $N$ 个点和 $M$ 条无向边的带权图，保证图中没有自环，但可能包含重边。

给出 $Q$ 次查询，每次查询给出 $K$ 条边 $B_1,B_2,\cdots ,B_K$，要求求出从节点 $1$ 到节点 $N$ 且这 $K$ 条边都至少经过一次的最短路（经过边的方向和顺序任意）。

赛时 Dijkstra 状态压缩 [TLE]

赛后

由于 $K$ 很小，因此我们可以枚举经过 $K$ 条边的顺序和方向，提前预处理出 $dist$ 数组即可直接计算出最短距离。

代码

#include <iostream>
#include <algorithm>
#include <cstring>
#define x first 
#define y second 

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<int, PII> PIP;

const int N = 405, M = 200005;

LL dist[N][N];
PIP edges[M];
int n, m, q, k;
int b[10], perm[10];

int main(){
    scanf("%d%d", &n, &m);
    memset(dist, 0x3f, sizeof dist);
    for (int i = 1; i <= n; i ++ ) dist[i][i] = 0;
    for (int i = 1; i <= m; i ++ ){
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        dist[u][v] = dist[v][u] = min(dist[u][v], 1ll * w);
        edges[i] = {w, {u, v}};
    }

    for (int k = 1; k <= n; k ++ )
        for (int i = 1; i <= n; i ++ )
            for (int j = 1; j <= n; j ++ )
                dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
    
    scanf("%d", &q);
    while (q -- ){
        scanf("%d", &k);
        for (int i = 1; i <= k; i ++ ) scanf("%d", &b[i]), perm[i] = i;

        LL ans = 0x3f3f3f3f3f3f3f3f;
        do {
            for (int state = 0; state < (1 << k); state ++ ){
                LL res = 0, last = 1;
                for (int i = 1; i <= k; i ++ ) {
                    PIP edge = edges[b[perm[i]]];
                    int from = edge.y.x, to = edge.y.y;
                    if ((state >> i - 1) & 1) swap(from, to);
                    res += dist[last][from] + edge.x;
                    last = to;
                }
                res += dist[last][n];
                ans = min(ans, res);
            }
        } while (next_permutation(perm + 1, perm + k + 1));

        printf("%lld\n", ans);
    }

    return 0;
}