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;
}
