旅游（CSUST省赛选拔赛2+状压dp+最短路）

题目链接：http://csustacm.com:4803/problem/1016

题目：

思路：状压dp+最短路，比赛的时候有想到状压dp，但是最短路部分写挫了，然后就卡死了，对不起出题人~dis[i][j]表示状态i下目的地为j时的最短路。

代码实现如下：

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef pair<ll, ll> pll;
 typedef pair<ll, int> pli;
 typedef pair<int, ll> pil;;
 typedef pair<int, int> pii;
 typedef unsigned long long ull;

 #define lson i<<1
 #define rson i<<1|1
 #define bug printf("*********\n");
 #define FIN freopen("D://code//in.txt", "r", stdin);
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = ;
 const int maxn = 2e5 + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f;

 int n, m, u, v, w;
 ll ans;
 int val[maxn], vis[][maxn];
 ll dis[][maxn];

 struct edge {
     int v, w;
     edge(int v, int w) : v(v), w(w) {}
 };

 struct node {
     int v, sta;
     ll w;
     node(int v, int sta, ll w) : v(v), sta(sta), w(w) {}
     bool operator < (const node& x) const {
         return w > x.w;
     }
 };

 vector<edge> G[maxn];

 void dij() {
     memset(vis ,, sizeof(vis));
     memset(dis, inf, sizeof(dis));
     priority_queue<node> q;
     q.push(node(, (<<val[]), ));
     dis[<<val[]][] = ;
     while(!q.empty()) {
         node e = q.top(); q.pop();
         int u = e.v, sta = e.sta;
         if(vis[sta][u]) continue;
         vis[sta][u] = ;
         for(int i = ; i < G[u].size(); i++) {
             int v = G[u][i].v, nw = sta | (<<val[v]);
             ll w = G[u][i].w;
             if(dis[nw][v] > dis[sta][u] + w) {
                 dis[nw][v] = dis[sta][u] + w;
                 q.push(node(v, nw, dis[nw][v]));
             }
         }
     }
 }

 int main() {
     //FIN;
     scanf("%d%d", &n, &m);
     for(int i = ; i <= n; i++) {
         scanf("%d", &val[i]);
     }
     for(int i = ; i <= m; i++) {
         scanf("%d%d%d", &u, &v, &w);
         G[u].push_back(edge(v, w));
         G[v].push_back(edge(u, w));
     }
     dij();
     ans = min(dis[][], dis[][]);
     printf("%lld\n", ans);
     return ;
 }