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：

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

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 + 5;
const int N = 1e3 + 5;
const int K = 25;
const int Sz_q = N * 64;
const int inf = 0x3f3f3f3f;
 
inline int read();
 
struct edge {
    int next, to, v;
    edge(int _n = 0, int _t = 0, int _v = 0) : next(_n), to(_t), v(_v) {}
} e[M << 1];
 
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[1024][N], g[32];
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 + 1) % Sz_q) {
        p = q[l], v[p] = 0, ++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] = 1;
                    if (dis[y] < dis[q[l]]) q[(l += Sz_q - 1) %= Sz_q] = y;
                    else q[++r %= Sz_q] = y;
                }
            }
    }
}
#undef y
 
int work() {
    static int S, s, i, res;
    for (S = 1; S < 1 << cnt; ++S) {
        for (i = 1; i <= n; ++i) {
            for (s = S & (S - 1); 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 = 1; i <= n; ++i)
        res = min(res, f[(1 << cnt) - 1][i]);
    return res;
}
 
int main() {
    int i, x, y, z, S, s, nowc;
    n = read(), m = read(), k = read();
    for (i = 1; i <= m; ++i) {
        x = read(), y = read(), z = read();
        Add_Edges(x, y, z);
    }
    for (i = 1; i <= k; ++i) p[i].get();
    sort(p + 1, p + k + 1);
    for (nowc = -1, c = 0, i = 1; i <= k; ++i) {
        if (p[i].c != nowc) nowc = p[i].c, ++c;
        p[i].c = c;
    }
     
    memset(g, 0x3f, sizeof(g));
    for (S = 1; S < 1 << c; ++S) {
        for (cnt = 0, i = 1; i <= k; ++i)
            if (S & (1 << p[i].c - 1)) ++cnt;
        memset(f, 0x3f, sizeof(f[0][0]) * N * (1 << cnt));
        for (cnt = 0, i = 1; i <= k; ++i)
            if (S & (1 << p[i].c - 1)) f[1 << cnt++][p[i].w] = 0;
        g[S] = work();
    }
    for (S = 1; S < 1 << c; ++S)
        for (s = S & (S - 1); s; (--s) &= S)
            g[S] = min(g[S], g[s] + g[S ^ s]);
    printf("%d
", g[(1 << c) - 1]);
    return 0;
}
 
inline int read() {
    static int x;
    static char ch;
    x = 0, ch = getchar();
    while (ch < '0' || '9' < ch)
        ch = getchar();
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x;
}