最短路 狀壓

設$f[i][s]$為到點$i$能殺的怪的狀態為$s$的最短路。用Dij刷分層圖最短路的時候轉移就好了。

程式碼：

#include<queue>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 205
#define M 3005
#define F inline
using namespace std;
struct edge{ int nxt,to,d,s; }ed[M<<
 
1];
struct node{ int x,d,s; };
int n,m,k,K,p,h[N],c[N],f[N][1<<14];
bool ff[N][1<<14];
priority_queue<node> q;
F char readc(){
	static char buf[100000],*l=buf,*r=buf;
	if (l==r) r=(l=buf)+fread(buf,1,100000,stdin);
	return l==r?EOF:*l++;
}
F int _read(){
	int x=0; char ch=readc();
	while (!isdigit
 
(ch)) ch=readc();
	while (isdigit(ch)) x=(x<<3)+(x<<1)+(ch^48),ch=readc();
	return x;
}
#define add(x,y,z,s) ed[++K]=(edge){h[x],y,z,s},h[x]=K
F bool operator < (node a,node b){ return a.d>b.d; }
F int Dij(){
	for (int i=1;i<=n;i++)
		for (int j=0;j<(1<<p);j++) f[i][j]=1e9
 
;
	for (f[1][0]=0,q.push((node){1,0,0});!q.empty();){
		int x=q.top().x,d=q.top().d,s=q.top().s; q.pop();
		if (x==n) return d; if (ff[x][s]) continue;
		ff[x][s]=true,s|=c[x];
		for (int i=h[x],v;i;i=ed[i].nxt)
			if ((ed[i].s&s)==ed[i].s&&d+ed[i].d<f[v=ed[i].to][s])
				f[v][s]=d+ed[i].d,q.push((node){v,f[v][s],s});
	}
	return -1;
}
int main(){
	n=_read(),m=_read(),p=_read(),k=_read();
	for (int i=1;i<=k;i++)
		for (int x=_read(),j=_read();j;j--) c[x]|=1<<_read()-1;
	for (int i=1,x,y,z,s;i<=m;i++){
		x=_read(),y=_read(),z=_read(),s=0;
		for (int j=_read();j;j--) s|=1<<_read()-1;
		add(x,y,z,s),add(y,x,z,s);
	}
	return printf("%d\n",Dij()),0;
}