[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

$1\le n\le8$

$2\le m\le30$

 1 #include <cstdio>

 2 #include <cstdlib>

 3 #include <cstring>

 4 #include <climits>

 5 #include <cctype>

 6 #include <cmath>

 7 #include <string>

 8 #include <sstream>

 9 #include <iostream>

10 #include <algorithm>

11 #include <iomanip>

12 using namespace std;

13 #include <queue>

14 #include <stack>

15 #include <vector>

16 #include <deque>

17 #include <set>

18 #include <map>

19 typedef long long LL;

20 typedef long double LD;

21 #define INFF 0x3f3f3f3f

22 #define INF 2139062143

23 #define pi acos(-1.0)

24 #define lson l, m, rt<<1

25 #define rson m+1, r, rt<<1|1

26 typedef pair<int, int> PI;

27 typedef pair<int, PI> PP;

28 #ifdef _WIN32

29 #define LLD "%I64d"

30 #else

31 #define LLD "%lld"

32 #endif

33 //#pragma comment(linker, "/STACK:1024000000,1024000000")

34 //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}

35 //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;}

36 //inline void print(LL x){printf(LLD, x);puts("");}

37 //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();}}

38 

39 int t[15];

40 int mp[35][35];

41 double dp[1<<10][35];  // 剩下的车票状态  现在在v的最小花费

42 int main()

43 {

44 #ifndef ONLINE_JUDGE

45     freopen("in.txt", "r", stdin);

46     freopen("out.txt", "w", stdout);

47 #endif

48     int n, m, p, a, b;

49     while(~scanf("%d%d%d%d%d", &n, &m, &p, &a, &b) && (n || m || p || a || b))

50     {

51         for(int i=0;i<n;i++)

52             scanf("%d", &t[i]);

53         memset(mp, -1, sizeof(mp));

54         while(p--)

55         {

56             int u, v, w;

57             scanf("%d%d%d", &u, &v, &w);

58             u--, v--;

59             if(mp[u][v]<0)

60                 mp[u][v]=mp[v][u]=w;

61             else

62                 mp[u][v]=mp[v][u]=min(mp[u][v], w);

63         }

64         memset(dp, 127, sizeof(dp));

65         dp[(1<<n)-1][a-1]=0;

66         double ans=INF;

67         for(int s=(1<<n)-1;s>=0;s--)

68         {

69             ans=min(ans, dp[s][b-1]);

70             for(int v=0;v<m;v++)

71                 for(int i=0;i<n;i++)

72                     if(s>>i & 1)

73                         for(int u=0;u<m;u++)

74                             if(mp[v][u]>=0)

75                                 dp[s & ~(1<<i)][u]=min(dp[s & ~(1<<i)][u], dp[s][v]+mp[v][u]*1.0/t[i]);

76    // 使用车票i , v->u

77         }

78         if(ans==INF)

79             printf("Impossible\n");

80         else

81             printf("%.3lf\n", ans);

82     }

83     return 0;

84 }

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