牛客练习赛76 D 魔物消灭计划 （斯坦纳树）

题目链接：https://ac.nowcoder.com/acm/contest/10845/D

斯坦纳树：联通若干特殊点的最小生成树

该题中因为同种宝石间传送无需花费代价，所以把同种宝石的城市缩成一个特殊点，然后求斯坦纳树即可

斯坦纳树求解：
设 (dp[i][S]) 表示以 (i) 为根，联通特殊点的状态为 (S) 花费的最小代价
转移分为两部分

1. (dp[i][S] = min(dp[i][S], dp[i][sub] + dp[i][S ^ sub])
   其中枚举子集使用以下技巧 (复杂度为 $O(3^n)):

for(int sub = S & (S - 1) ; sub ; sub = S & (sub - 1))

2. (dp[i][S] = min(dp[i][S], dp[j][S] + w[i][j]))
   第二部分转移就是三角形不等式，dijstra即可

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;

#define mk make_pair

const int maxn = 610;

int n, m, k, x, y; 
int a[maxn], id[maxn];
int dp[maxn][(1 << 10) + 10];

int h[maxn], cnt;
struct E{
	int to, next, cost;
}e[maxn << 1];
void add(int u, int v, int w){
	e[++cnt].to = v;
	e[cnt].cost = w;
	e[cnt].next = h[u];
	h[u] = cnt;
}

priority_queue<P, vector<P>, greater<P> > q;
int vis[maxn];

void dij(int S){
	memset(vis, 0, sizeof(vis));
	while(!q.empty()){
		P p = q.top(); q.pop();
		int u = p.second;
		
		if(vis[u]) continue;
		vis[u] = 1;
		
		for(int i = h[u] ; i != -1 ; i = e[i].next){
			int w = e[i].cost, v = e[i].to;
			if(dp[v][S] > dp[u][S] + w){
				dp[v][S] = dp[u][S] + w;
				q.push(mk(dp[v][S], v));
			}
		}
	}
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
	memset(h, -1, sizeof(h));
	
	n = read(), m = read(), k = read(), x = read(), y = read();
	id[x] = k + 1, id[y] = k + 2; k += 2;
	
	int tot = k;
	for(int i = 1 ; i <= n ; ++i){
		a[i] = read();
		if(i == x || i == y) continue;
		if(a[i]) id[i] = a[i];
		else id[i] = ++tot;
	}
	
	n = tot;
	
	int u, v, w;
	for(int i = 1 ; i <= m ; ++i){
		u = read(), v = read(), w = read();
		add(id[u], id[v], w), add(id[v], id[u], w);
	}
	
	memset(dp, 0x3f, sizeof(dp));
	
	for(int i = 1 ; i <= k ; ++i) dp[i][1 << (i - 1)] = 0;
	
	for(int S = 1 ; S < (1 << k) ; ++S){
		for(int i = 1 ; i <= n ; ++i){
			for(int sub = S & (S - 1) ; sub ; sub = S & (sub - 1)){
				dp[i][S] = min(dp[i][S], dp[i][sub] + dp[i][S ^ sub]);
			}
			if(dp[i][S] != 0x3f3f3f3f) q.push(mk(dp[i][S], i));
		}
		dij(S);
	}
	
	printf("%d
", dp[1][(1 << k) - 1]);
	
	return 0;
}