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：

 1 #include <set>
 2 #include <map>
 3 #include <queue>
 4 #include <stack>
 5 #include <cmath>
 6 #include <bitset>
 7 #include <cstdio>
 8 #include <string>
 9 #include <vector>
10 #include <cstdlib>
11 #include <cstring>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 
16 typedef long long ll;
17 typedef pair<ll, ll> pll;
18 typedef pair<ll, int> pli;
19 typedef pair<int, ll> pil;;
20 typedef pair<int, int> pii;
21 typedef unsigned long long ull;
22 
23 #define lson i<<1
24 #define rson i<<1|1
25 #define bug printf("*********
");
26 #define FIN freopen("D://code//in.txt", "r", stdin);
27 #define debug(x) cout<<"["<<x<<"]" <<endl;
28 #define IO ios::sync_with_stdio(false),cin.tie(0);
29 
30 const double eps = 1e-8;
31 const int mod = 10007;
32 const int maxn = 2e5 + 7;
33 const double pi = acos(-1);
34 const int inf = 0x3f3f3f3f;
35 const ll INF = 0x3f3f3f3f3f3f3f;
36 
37 int n, m, u, v, w;
38 ll ans;
39 int val[maxn], vis[10][maxn];
40 ll dis[10][maxn];
41 
42 struct edge {
43     int v, w;
44     edge(int v, int w) : v(v), w(w) {}
45 };
46 
47 struct node {
48     int v, sta;
49     ll w;
50     node(int v, int sta, ll w) : v(v), sta(sta), w(w) {}
51     bool operator < (const node& x) const {
52         return w > x.w;
53     }
54 };
55 
56 vector<edge> G[maxn];
57 
58 void dij() {
59     memset(vis ,0, sizeof(vis));
60     memset(dis, inf, sizeof(dis));
61     priority_queue<node> q;
62     q.push(node(1, (1<<val[1]), 0));
63     dis[1<<val[1]][1] = 0;
64     while(!q.empty()) {
65         node e = q.top(); q.pop();
66         int u = e.v, sta = e.sta;
67         if(vis[sta][u]) continue;
68         vis[sta][u] = 1;
69         for(int i = 0; i < G[u].size(); i++) {
70             int v = G[u][i].v, nw = sta | (1<<val[v]);
71             ll w = G[u][i].w;
72             if(dis[nw][v] > dis[sta][u] + w) {
73                 dis[nw][v] = dis[sta][u] + w;
74                 q.push(node(v, nw, dis[nw][v]));
75             }
76         }
77     }
78 }
79 
80 int main() {
81     //FIN;
82     scanf("%d%d", &n, &m);
83     for(int i = 1; i <= n; i++) {
84         scanf("%d", &val[i]);
85     }
86     for(int i = 1; i <= m; i++) {
87         scanf("%d%d%d", &u, &v, &w);
88         G[u].push_back(edge(v, w));
89         G[v].push_back(edge(u, w));
90     }
91     dij();
92     ans = min(dis[6][1], dis[7][1]);
93     printf("%lld
", ans);
94     return 0;
95 }