Abandoning Roads CodeForces

大意: 给定无向图, 边权只有两种, 对于每个点$x$, 输出所有最小生成树中, 点$1$到$x$的最短距离.

先将边权为$a$的边合并, 考虑添加边权为$b$的边.

每条路径只能经过每个连通块一次, 直接状压的话有$O(n2^n)$个状态.

但是注意到点数不超过$3$的连通块内部最短路不超过$2a$, 所以求最短路时一定只经过$1$次, 所以可以不考虑.

这样总状态就为$O(n2^{frac{n}{4}})$.

#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 1e7+10;
int n, m, tot, a, b, fa[555], sz[555], id[555];
struct _ {int to,w;};
vector<_> g[555];
int Find(int x) {return fa[x]?fa[x]=Find(fa[x]):x;}
void add(int x, int y) {
	x=Find(x),y=Find(y);
	if (x!=y) fa[x]=y,sz[y]+=sz[x];
}
struct __ {
	int id, w;
	bool operator < (const __ &rhs) const {
		return w>rhs.w;
	}
};
priority_queue<__> q;
int dis[N], cnt;
int ID(int x, int y) {
	return x*n+y;
}
pii get(int s) {
	return pii((s-1)/n,(s-1)%n+1);
}

int main() {
	scanf("%d%d%d%d", &n, &m, &a, &b);
	REP(i,1,n) sz[i]=1;
	REP(i,1,m) {
		int u, v, c;
		scanf("%d%d%d", &u, &v, &c);
		g[u].pb({v,c}),g[v].pb({u,c});
		if (c==a) add(u,v);
	}
	REP(i,1,n) if (i==Find(i)&&sz[i]>3) {
		REP(j,1,n) if (Find(j)==i) id[j] = 1<<tot;
		++tot;
	}
	int mx = (1<<tot)-1;
	memset(dis,0x3f,sizeof dis);
	q.push({ID(id[1],1),dis[ID(id[1],1)]=0});
	while (q.size()) {
		__ t = q.top(); q.pop();
		int u = t.id, w = t.w;
		if (dis[u]!=w) continue;
		for (auto &e:g[get(u).y]) {
			int v = ID(get(u).x|id[e.to],e.to);
			if (e.w==a) {
				if (dis[v]>w+a) q.push({v,dis[v]=w+a});
			}
			else if (Find(get(u).y)!=Find(e.to)&&!(get(u).x&id[e.to])) {
				if (dis[v]>w+b) q.push({v,dis[v]=w+b});
			}
		}
	}
	REP(i,1,n) {
		int ans = INF;
		REP(j,0,mx) ans = min(ans, dis[ID(j,i)]);
		printf("%d ", ans);
	} hr;
}