深搜加剪纸可A（O(玄学) 1274ms）
——代码
1 #include <cmath> 2 #include <cstdio> 3 #include <iostream> 4 5 int n; 6 double ans = ~(1 << 31), a[16], b[16]; 7 bool vis[16]; 8 9 inline double min(double x, double y) 10 { 11 return x < y ? x : y; 12 } 13 14 inline double query(int x, int y) 15 { 16 return sqrt((a[x] - a[y]) * (a[x] - a[y]) + (b[x] - b[y]) * (b[x] - b[y])); 17 } 18 19 inline void dfs(int now, double sum, int k) 20 { 21 if(k == n + 1) 22 { 23 ans = min(ans, sum); 24 return; 25 } 26 if(sum > ans) return; 27 for(int i = 1; i <= n; i++) 28 if(!vis[i]) 29 { 30 vis[i] = 1; 31 dfs(i, sum + query(now, i), k + 1); 32 vis[i] = 0; 33 } 34 } 35 36 int main() 37 { 38 scanf("%d", &n); 39 for(int i = 1; i <= n; i++) scanf("%lf %lf", &a[i], &b[i]); 40 dfs(0, 0, 1); 41 printf("%.2lf ", ans); 42 return 0; 43 }
然而状压DP，稳一手（据说O(n2*2n) 73ms）
采用记忆化搜索。
设f[i][S]为已经走过的点的集合为S，当前停留在点i的最短距离
f[i][S] = f[j][S - i] + dis(i, j) (i, j ∈ S && i != j)
——代码
1 #include <cmath> 2 #include <cstdio> 3 4 const int INF = 1e9; 5 int n; 6 double ans, a[16], b[16], f[16][1 << 16]; 7 8 inline double dist(int i, int j) 9 { 10 return sqrt((a[i] - a[j]) * (a[i] - a[j]) + (b[i] - b[j]) * (b[i] - b[j])); 11 } 12 13 inline double min(double x, double y) 14 { 15 return x < y ? x : y; 16 } 17 18 inline int istrue(int x, int S) 19 { 20 return (1 << x - 1) & S; 21 } 22 23 inline int set0(int x, int S) 24 { 25 return (~(1 << x - 1)) & S; 26 } 27 28 inline void dfs(int now, int S) 29 { 30 if(f[now][S] != -1) return; 31 f[now][S] = INF; 32 for(int i = 1; i <= n; i++) 33 { 34 if(i == now) continue; 35 if(!istrue(i, S)) continue; 36 dfs(i, set0(now, S)); 37 f[now][S] = min(f[now][S], f[i][set0(now, S)] + dist(i, now)); 38 } 39 } 40 41 int main() 42 { 43 int i, j; 44 scanf("%d", &n); 45 for(i = 1; i <= n; i++) scanf("%lf %lf", &a[i], &b[i]); 46 for(i = 1; i <= n; i++) 47 for(j = 1; j < (1 << n); j++) 48 f[i][j] = -1; 49 for(i = 1; i <= n; i++) f[i][1 << i - 1] = dist(0, i); 50 for(i = 1; i <= n; i++) dfs(i, (1 << n) - 1); 51 ans = INF; 52 for(i = 1; i <= n; i++) ans = min(ans, f[i][(1 << n) - 1]); 53 printf("%.2lf ", ans); 54 return 0; 55 }
用递推更简便（68ms）
——代码
1 #include <cmath> 2 #include <cstdio> 3 #include <cstring> 4 5 const int INF = 1e9; 6 int n, m; 7 double ans, a[16], b[16], f[16][1 << 16]; 8 9 inline double dist(int i, int j) 10 { 11 return sqrt((a[i] - a[j]) * (a[i] - a[j]) + (b[i] - b[j]) * (b[i] - b[j])); 12 } 13 14 inline double min(double x, double y) 15 { 16 return x < y ? x : y; 17 } 18 19 int main() 20 { 21 int i, j, S, x, y; 22 scanf("%d", &n); 23 m = (1 << n) - 1; 24 for(i = 1; i <= n; i++) scanf("%lf %lf", &a[i], &b[i]); 25 memset(f, 0x7f, sizeof(f)); 26 for(i = 1; i <= n; i++) f[i][1 << i - 1] = dist(0, i); 27 for(S = 1; S <= m; S++) 28 for(i = 1; i <= n; i++) 29 { 30 if(!((1 << i - 1) & S)) continue; 31 for(j = 1; j <= n; j++) 32 { 33 if(i == j) continue; 34 if(!((1 << j - 1) & S)) continue; 35 f[i][S] = min(f[i][S], f[j][(1 << i - 1) ^ S] + dist(j, i)); 36 } 37 } 38 ans = INF; 39 for(i = 1; i <= n; i++) ans = min(ans, f[i][m]); 40 printf("%.2lf ", ans); 41 return 0; 42 }