BZOJ4006 [JLOI2015]管道连接

裸的状压DP

令$f_S$表示包含颜色集合S的最小斯坦纳生成森林的值，于是有：

$$f_S=\min\{f_S,f_s+f_{S-s}|s\subset S\}$$

然后嘛。。。还是裸的斯坦纳树搞搞。。。又是个状压【摔！

貌似会TLE的说【额。。。

然后PoPoQQQ大爷分析了一番，说，大概1E的复杂度，不会T！

好，那就不会好了！(也太不求上进了吧)

 /**************************************************************
     Problem: 4006
     User: rausen
     Language: C++
     Result: Accepted
     Time:7272 ms
     Memory:5160 kb
 ****************************************************************/

 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 using namespace std;
 const int M = 3e3 + ;
 const int N = 1e3 + ;
 const int K = ;
 const int Sz_q = N * ;
 const int inf = 0x3f3f3f3f;

 inline int read();

 struct edge {
     int next, to, v;
     edge(int _n = , int _t = , int _v = ) : next(_n), to(_t), v(_v) {}
 } e[M << ];

 struct point {
     int c, w;

     inline void get() {
         c = read(), w = read();
     }

     inline bool operator < (const point &p) const {
         return c < p.c;
     }
 } p[K];

 int n, m, k, c, cnt;
 int first[N], tot;
 int f[][N], g[];
 int l, r, q[Sz_q], v[N];

 inline void Add_Edges(int x, int y, int z) {
     e[++tot] = edge(first[x], y, z), first[x] = tot;
     e[++tot] = edge(first[y], x, z), first[y] = tot;
 }

 #define y e[x].to
 void spfa(int *dis) {
     static int x, p;
     while (l != (r + ) % Sz_q) {
         p = q[l], v[p] = , ++l %= Sz_q;
         for (x = first[p]; x; x = e[x].next)
             if (dis[p] + e[x].v < dis[y]) {
                 dis[y] = dis[p] + e[x].v;
                 if (!v[y]) {
                     v[y] = ;
                     if (dis[y] < dis[q[l]]) q[(l += Sz_q - ) %= Sz_q] = y;
                     else q[++r %= Sz_q] = y;
                 }
             }
     }
 }
 #undef y

 int work() {
     static int S, s, i, res;
     for (S = ; S <  << cnt; ++S) {
         for (i = ; i <= n; ++i) {
             for (s = S & (S - ); s; (--s) &= S)
                 f[S][i] = min(f[S][i], f[s][i] + f[S ^ s][i]);
             if (f[S][i] != inf) q[++r %= Sz_q] = i;
         }
         spfa(f[S]);
     }
     for (res = inf, i = ; i <= n; ++i)
         res = min(res, f[( << cnt) - ][i]);
     return res;
 }

 int main() {
     int i, x, y, z, S, s, nowc;
     n = read(), m = read(), k = read();
     for (i = ; i <= m; ++i) {
         x = read(), y = read(), z = read();
         Add_Edges(x, y, z);
     }
     for (i = ; i <= k; ++i) p[i].get();
     sort(p + , p + k + );
     for (nowc = -, c = , i = ; i <= k; ++i) {
         if (p[i].c != nowc) nowc = p[i].c, ++c;
         p[i].c = c;
     }

     memset(g, 0x3f, sizeof(g));
     for (S = ; S <  << c; ++S) {
         for (cnt = , i = ; i <= k; ++i)
             if (S & ( << p[i].c - )) ++cnt;
         memset(f, 0x3f, sizeof(f[][]) * N * ( << cnt));
         for (cnt = , i = ; i <= k; ++i)
             if (S & ( << p[i].c - )) f[ << cnt++][p[i].w] = ;
         g[S] = work();
     }
     for (S = ; S <  << c; ++S)
         for (s = S & (S - ); s; (--s) &= S)
             g[S] = min(g[S], g[s] + g[S ^ s]);
     printf("%d\n", g[( << c) - ]);
     return ;
 }

 inline int read() {
     static int x;
     static char ch;
     x = , ch = getchar();
     while (ch < '' || '' < ch)
         ch = getchar();
     while ('' <= ch && ch <= '') {
         x = x *  + ch - '';
         ch = getchar();
     }
     return x;
 }