[状压dp]POJ2686 Traveling by Stagecoach

题意: m个城市, n张车票, 每张车票$t_i$匹马, 每张车票可以沿某条道路到相邻城市, 花费是路的长度除以马的数量. 求a到b的最小花费, 不能到达输出Impossible

$1\le n\le8$

$2\le m\le30$

 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cmath>
 #include <string>
 #include <sstream>
 #include <iostream>
 #include <algorithm>
 #include <iomanip>
 using namespace std;
 #include <queue>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <set>
 #include <map>
 typedef long long LL;
 typedef long double LD;
 #define INFF 0x3f3f3f3f
 #define INF 2139062143
 #define pi acos(-1.0)
 #define lson l, m, rt<<1
 #define rson m+1, r, rt<<1|1
 typedef pair<int, int> PI;
 typedef pair<int, PI> PP;
 #ifdef _WIN32
 #define LLD "%I64d"
 #else
 #define LLD "%lld"
 #endif
 //#pragma comment(linker, "/STACK:1024000000,1024000000")
 //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}
 //inline int read(){char ch=' ';int ans=0;while(ch<'0' || ch>'9')ch=getchar();while(ch<='9' && ch>='0'){ans=ans*10+ch-'0';ch=getchar();}return ans;}
 //inline void print(LL x){printf(LLD, x);puts("");}
 //inline void read(int &x){char c = getchar();while(c < '0') c = getchar();x = c - '0'; c = getchar();while(c >= '0'){x = x * 10 + (c - '0'); c = getchar();}}

 int t[];
 int mp[][];
 double dp[<<][];  // 剩下的车票状态  现在在v的最小花费
 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w", stdout);
 #endif
     int n, m, p, a, b;
     while(~scanf("%d%d%d%d%d", &n, &m, &p, &a, &b) && (n || m || p || a || b))
     {
         for(int i=;i<n;i++)
             scanf("%d", &t[i]);
         memset(mp, -, sizeof(mp));
         while(p--)
         {
             int u, v, w;
             scanf("%d%d%d", &u, &v, &w);
             u--, v--;
             if(mp[u][v]<)
                 mp[u][v]=mp[v][u]=w;
             else
                 mp[u][v]=mp[v][u]=min(mp[u][v], w);
         }
         memset(dp, , sizeof(dp));
         dp[(<<n)-][a-]=;
         double ans=INF;
         for(int s=(<<n)-;s>=;s--)
         {
             ans=min(ans, dp[s][b-]);
             for(int v=;v<m;v++)
                 for(int i=;i<n;i++)
                     if(s>>i & )
                         for(int u=;u<m;u++)
                             if(mp[v][u]>=)
                                 dp[s & ~(<<i)][u]=min(dp[s & ~(<<i)][u], dp[s][v]+mp[v][u]*1.0/t[i]);
    // 使用车票i , v->u
         }
         if(ans==INF)
             printf("Impossible\n");
         else
             printf("%.3lf\n", ans);
     }
     return ;
 }