【状态压缩 搜索】SSL_1150 密室

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题意

有 $N$个房间， $M$个传送门， $K$种钥匙，每个房间有一些钥匙，每个传送门可以从一个房间传到另一个房间，需要一定的钥匙才能使用传送门，求出经过最少的传送门能从 $1$号房间到达 $N$号房间。

思路

压缩钥匙的状态，然后 $BFS$就行了。

代码

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

struct node{
	int to, k, next;
}e[6001];
struct state{
	int w, k;
};
int N, M, K, ans, x;
int key[5001], list[5001], d[2048][5001];

void bfs() {
	queue<state> q;
	q.push((state){1, key[1]});
	memset(d, -1, sizeof(d));
	d[key[1]][1] = 0;
	while (q.size()) {
		state head = q.front();
		q.pop();
		for (int i = list[head.w]; i; i = e[i].next) {
			int s = head.k, y = e[i].to;
			if ((s & e[i].k) == e[i].k && d[s | key[y]][y] == -1) {//判断分层
				q.push((state){y, s | key[y]});
				d[s | key[y]][y] = d[s][head.w] + 1;
			}
		}
		if (head.w == N) ans = min(ans, d[head.k][N]);
	}
}

int main() {
	scanf("%d %d %d", &N, &M, &K);
	for (int i = 1; i <= N; i++) {
		for (int j = 1; j <= K; j++) {
			scanf("%d", &x);
			key[i] = key[i] << 1 | x;
		}
	}
	for (int i = 1; i <= M; i++) {
		scanf("%d %d", &x, &e[i].to);
		e[i].next = list[x];
		list[x] = i;
		for (int j = 1; j <= K; j++) {
			scanf("%d", &x);
			e[i].k = e[i].k << 1 | x;
		}
	}
	ans = 2147483647;
	bfs();
	if (ans == 2147483647) printf("No Solution");
	else printf("%d", ans);
}