[luoguP1433] 吃奶酪（DP || Dfs）

传送门

深搜加剪纸可A（O(玄学) 1274ms）

——代码

 #include <cmath>
 #include <cstdio>
 #include <iostream>

 int n;
 double ans = ~( << ), a[], b[];
 bool vis[];

 inline double min(double x, double y)
 {
     return x < y ? x : y;
 }

 inline double query(int x, int y)
 {
     return sqrt((a[x] - a[y]) * (a[x] - a[y]) + (b[x] - b[y]) * (b[x] - b[y]));
 }

 inline void dfs(int now, double sum, int k)
 {
     if(k == n + )
     {
         ans = min(ans, sum);
         return;
     }
     if(sum > ans) return;
     for(int i = ; i <= n; i++)
         if(!vis[i])
         {
             vis[i] = ;
             dfs(i, sum + query(now, i), k + );
             vis[i] = ;
         }
 }

 int main()
 {
     scanf("%d", &n);
     for(int i = ; i <= n; i++) scanf("%lf %lf", &a[i], &b[i]);
     dfs(, , );
     printf("%.2lf\n", ans);
     return ;
 }

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

 #include <cmath>
 #include <cstdio>

 const int INF = 1e9;
 int n;
 double ans, a[], b[], f[][ << ];

 inline double dist(int i, int j)
 {
     return sqrt((a[i] - a[j]) * (a[i] - a[j]) + (b[i] - b[j]) * (b[i] - b[j]));
 }

 inline double min(double x, double y)
 {
     return x < y ? x : y;
 }

 inline int istrue(int x, int S)
 {
     return ( << x - ) & S;
 }

 inline int set0(int x, int S)
 {
     return (~( << x - )) & S;
 }

 inline void dfs(int now, int S)
 {
     if(f[now][S] != -) return;
     f[now][S] = INF;
     for(int i = ; i <= n; i++)
     {
         if(i == now) continue;
         if(!istrue(i, S)) continue;
         dfs(i, set0(now, S));
         f[now][S] = min(f[now][S], f[i][set0(now, S)] + dist(i, now));
     }
 }

 int main()
 {
     int i, j;
     scanf("%d", &n);
     for(i = ; i <= n; i++) scanf("%lf %lf", &a[i], &b[i]);
     for(i = ; i <= n; i++)
         for(j = ; j < ( << n); j++)
             f[i][j] = -;
     for(i = ; i <= n; i++) f[i][ << i - ] = dist(, i);
     for(i = ; i <= n; i++) dfs(i, ( << n) - );
     ans = INF;
     for(i = ; i <= n; i++) ans = min(ans, f[i][( << n) - ]);
     printf("%.2lf\n", ans);
     return ;
 }

用递推更简便（68ms）

——代码

 #include <cmath>
 #include <cstdio>
 #include <cstring>

 const int INF = 1e9;
 int n, m;
 double ans, a[], b[], f[][ << ];

 inline double dist(int i, int j)
 {
     return sqrt((a[i] - a[j]) * (a[i] - a[j]) + (b[i] - b[j]) * (b[i] - b[j]));
 }

 inline double min(double x, double y)
 {
     return x < y ? x : y;
 }

 int main()
 {
     int i, j, S, x, y;
     scanf("%d", &n);
     m = ( << n) - ;
     for(i = ; i <= n; i++) scanf("%lf %lf", &a[i], &b[i]);
     memset(f, 0x7f, sizeof(f));
     for(i = ; i <= n; i++) f[i][ << i - ] = dist(, i);
     for(S = ; S <= m; S++)
         for(i = ; i <= n; i++)
         {
             if(!(( << i - ) & S)) continue;
             for(j = ; j <= n; j++)
             {
                 if(i == j) continue;
                 if(!(( << j - ) & S)) continue;
                 f[i][S] = min(f[i][S], f[j][( << i - ) ^ S] + dist(j, i));
             }
         }
     ans = INF;
     for(i = ; i <= n; i++) ans = min(ans, f[i][m]);
     printf("%.2lf\n", ans);
     return ;
 }