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
$1le nle8$
$2le mle30$
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 "); 80 else 81 printf("%.3lf ", ans); 82 } 83 return 0; 84 }