【模板】最小斯坦纳树
作者：互联网
模型，给定 \(n\) 个点 \(m\) 条边的无向图，和 \(k\) 个关键点。选出边权和最小的一些边使得 \(k\) 个点连通。
因为选出的边一定是一棵树，所以称为最小斯坦纳树。
直接状压 \(f[i][S]\) 表示以 \(i\) 为根，与 \(i\) 联通的关键点集合为 \(S\)。
我们可以枚举子集直接转移 \(f[i][S] = \min\limits_{T\subset S}\{f[i][T] + f[i][S\oplus T]\}\)。
也可以再外接一条边，\(f[i][s] + e_{i,j} \to f[j][s]\)。
第一种转移的阶段很清晰，直接枚举子集即可。第二种转移有后效性，我们对每层 \(s\) 跑最短路即可。
时间复杂度 \(\mathcal{O}(n3^k + 2^km\log m)\)。
/*
Author : SharpnessV
Right Output ! & Accepted !
*/
#include<bits/stdc++.h>
//#define int long long
#define rep(i, a, b) for(int i = (a);i <= (b);i++)
#define pre(i, a, b) for(int i = (a);i >= (b);i--)
#define rp(i, a) for(int i = 1; i <= (a); i++)
#define pr(i, a) for(int i = (a); i >= 1; i--)
#define go(i, x) for(auto i : x)
#define mp make_pair
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define ze(p) memset(p, 0, sizeof(p))
#define mem(p, x) memset(p, x, sizeof(p))
#define YES puts("YES")
#define NO puts("NO")
#define Yes puts("Yes")
#define No puts("No")
#define si(x) (int)(x).size()
#define db cerr
#define pc putchar
#define gc getchar
#define el putchar('\n')
using namespace std;
const double eps = 1e-15, pi = 3.1415926535897932385;
typedef long long LL;
typedef pair<int,int> Pr;
const int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}, inf = 0x7fffffff;
//Author : SharpnessV
//#define main Solution
//char buf[1<<22],*p1=buf,*p2=buf;
//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
inline int read(){
int x = 0;bool f = 1;char ch = gc();
while(!isdigit(ch))f = ('-' == ch ? 0 : 1), ch = gc();
while(isdigit(ch))x = (x << 1) + (x << 3) + (ch ^ 48), ch = gc();
if(f)return x;return -x;
}
inline LL Read(){
LL x = 0;bool f = 1;char ch = gc();
while(!isdigit(ch))f = ('-' == ch ? 0 : 1), ch = gc();
while(isdigit(ch))x = (x << 1) + (x << 3) + (ch ^ 48), ch = gc();
if(f)return x;return -x;
}
int gcd(int x,int y){return y ? gcd(y, x % y) : x;}
int lcm(int x,int y){return x / gcd(x, y) * y;}
#define P 1000000007
//#define P 998244353
#define bas 229
inline void ad(int &x, int y){x += y; if(x >= P) x -= P;}
inline void su(int &x, int y){x -= y; if(x < 0) x += P;}
inline void cmn(int &x,int y){if(y < x) x = y;}
inline void cmx(int &x,int y){if(y > x) x = y;}
inline void cmn(LL &x, LL y){if(y < x) x = y;}
inline void cmx(LL &x, LL y){if(y > x) x = y;}
int Pow(int x, int y){
if(y < 0)return Pow(Pow(x, P - 2), -y);
int now = 1 ;
for(;y;y >>= 1, x = 1LL * x * x % P)if(y & 1) now = 1LL * now * x % P;
return now;
}
/*******************************************************************************************************************/
/* */
/*******************************************************************************************************************/
#define N 105
int n, m, k, f[N][1 << 10], v[N];
vector<Pr>e[N];
priority_queue<Pr>q;
void dij(int s){
while(!q.empty())q.pop();
rp(i, n)q.push(mp(-f[i][s], i));
memset(v, 0, sizeof(v));
while(!q.empty()){
int x = q.top().se;
q.pop(), v[x] = 1;
go(y, e[x])if(f[x][s] + y.se < f[y.fi][s])
f[y.fi][s] = f[x][s] + y.se, q.push(mp(-f[y.fi][s], y.fi));
while(!q.empty() && v[q.top().se])q.pop();
}
}
int main(){
//int T = read();while(T--)solve();
n = read(), m = read(), k = read();
rp(i, m){
int x = read(), y = read(), z = read();
e[x].pb(mp(y, z)), e[y].pb(mp(x, z));
}
memset(f, 0x3f, sizeof(f));
rp(i, k){
int x = read();
f[x][1 << (i - 1)] = 0;
}
rp(i, (1 << k) - 1){
rp(x, n)
for(int j = (i - 1) & i; j ;j = (j - 1) & i)cmn(f[x][i], f[x][j] + f[x][i ^ j]);
dij(i);
}
int ans = inf;
rp(i, n)cmn(ans, f[i][(1 << k) - 1]);
printf("%d\n", ans);
return 0;
}
标签：int,void,最小,斯坦纳,read,mp,fi,模板,define 来源： https://www.cnblogs.com/SharpnessV/p/15203862.html