bzoj1402 Ticket to Ride 斯坦纳树 + 状压dp

给定(n)个点，(m)条边的带权无向图

选出一些边，使得(4)对点之间可达，询问权值最小为多少

(n leqslant 30, m leqslant 1000)

---

首先看数据范围，(4)对点，也就是(8)个点，很小

上斯坦纳树（局部最小生成树）

然而好像题目并不是斯坦纳树，可能是一些树拼到一起

那么就再做一个状压(dp)即可

复杂度(O(3^8 * n + 2^8 * nm + 2^{12} * n))

---

#include <map>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int sid = 55;
const int eid = (1 << 9) - 1;
inline void cmin(int &a, int b) { if(a > b) a = b; }

int n, m, nc;
int U[sid], V[sid], cl[sid], ip[sid];
int f[sid][eid], E[sid][sid];
string s[55], sa, sb;
map <string, int> id;

int inq[sid];
queue <int> q;
void spfa(int S) {
	memset(inq, 0, sizeof(inq));
	for(int i = 1; i <= n; i ++) q.push(i);
	while(!q.empty()) {
		int id = q.front(); q.pop(); inq[id] = 0;
		for(int j = 1; j <= n; j ++)
			if(f[j][S] > f[id][S] + E[id][j]) {
				f[j][S] = f[id][S] + E[id][j];
				if(!inq[j]) q.push(j), inq[j] = 1;
			}
	}
}

void wish1() {
	memset(f, 56, sizeof(f));
	for(int i = 1; i <= n; i ++)
		if(cl[i]) {
			nc ++;
			ip[i] = nc;
			f[i][(1 << nc - 1)] = 0;
		}
		
	for(int S = 0; S <= (1 << nc) - 1; S ++) {
		for(int i = 1; i <= n; i ++)
			for(int T = S; T; T = (T - 1) & S)
				cmin(f[i][S], f[i][T] + f[i][S ^ T]);
		spfa(S);
	}
}

int g[eid];
void wish2() {
	memset(g, 56, sizeof(g)); g[0] = 0;
	for(int s = 0; s <= (1 << 4) - 1; s ++)
	for(int i = 1; i <= n; i ++)
	for(int S = 0; S <= (1 << nc) - 1; S ++) {
		int T = 0;
		for(int k = 1; k <= 4; k ++)
			if((S & (1 << ip[U[k]] - 1)) && (S & (1 << ip[V[k]] - 1)))
				T |= (1 << k - 1);
		cmin(g[s | T], g[s] + f[i][S]);
	}
	printf("%d
", g[(1 << 4) - 1]);
}

int main() {
	freopen("bzoj1402.in", "r", stdin);
	freopen("bzoj1402.out", "w", stdout);
	cin >> n >> m;
	for(int i = 1; i <= n; i ++) {
		cin >> s[i];
		id[s[i]] = i;
	}
	memset(E, 56, sizeof(E));
	for(int i = 1; i <= m; i ++) {
		int u, v, w;
		cin >> sa >> sb >> w;
		u = id[sa]; v = id[sb];
		cmin(E[u][v], w); E[v][u] = E[u][v];
	}
	for(int i = 1; i <= 4; i ++) {
		int u, v;
		cin >> sa >> sb;
		u = id[sa]; v = id[sb];
		U[i] = u; V[i] = v;
		cl[u] = cl[v] = 1;
	}
	wish1(); wish2();
	return 0;
}