Solution -「CF 575G」Run for beer

$\mathcal{Description}$

  Link.

  给定 $n$ 个点 $m$ 条边的无向图，边有边权，一个人初始速度为 $1$，每走一条边速度 $\div10$，求从 $1$ 走到 $n$ 的最小耗时。

  $n,m\le10^5$，$0\le\text{边权}\le9$。

$\mathcal{Solution}$

  直观地，路径长度即为把经过的边权从低位到高位写成的十进制数。

  首先排除前导 $0$——把从终点出发，仅走边权为 $0$ 的边可达的结点全部与终点缩点。此时的最短路需要保证路径条数最少的前提下保证字典序最小。BFS 分层，维护当前层外围最优的一堆结点，用它们向下层扩展直到到达起点。

$\mathcal{Code}$

#include <queue>
#include <cstdio>
#include <vector>

const int MAXN = 2e5, MAXM = 2e5;
int n, m, ecnt, head[MAXN + 5], d[MAXN + 5], suf[MAXN + 5];
bool vis[MAXN + 5];
std::vector<int> curp, nxtp;
std::queue<int> que;

struct Edge { int to, cst, nxt; } graph[MAXM * 2 + 5];

inline void link ( const int s, const int t, const int c ) {
	graph[++ ecnt] = { t, c, head[s] };
	head[s] = ecnt;
}

inline void initReach () {
	for ( int i = 1; i <= n; ++ i ) d[i] = -1;
	d[1] = 0, que.push ( 1 );
	for ( int u; ! que.empty (); ) {
		u = que.front (), que.pop ();
		for ( int i = head[u], v; i; i = graph[i].nxt ) {
			if ( ! ~ d[v = graph[i].to] ) {
				d[v] = d[u] + 1, que.push ( v );
			}
		}
	}
}

inline int zeroReach () {
	int mind = d[n];
	curp.push_back ( n ), vis[n] = true;
	for ( int cur = 0; cur ^ curp.size (); ++ cur ) {
		int u = curp[cur];
		for ( int i = head[u], v; i; i = graph[i].nxt ) {
			if ( ! vis[v = graph[i].to] && ! graph[i].cst ) {
				curp.push_back ( v ), vis[v] = true, suf[v] = u;
				if ( mind > d[v] ) mind = d[v];
			}
		}
	}
	return mind;
}

int main () {
	scanf ( "%d %d", &n, &m );
	for ( int i = 1, u, v, w; i <= m; ++ i ) {
		scanf ( "%d %d %d", &u, &v, &w ), ++ u, ++ v;
		link ( u, v, w ), link ( v, u, w );
	}
	initReach ();
	int dist = zeroReach ();
	bool zero = true;
	for ( int l = dist; l; -- l ) {
		int dig = 10;
		for ( int u: curp ) {
			for ( int i = head[u], v; i; i = graph[i].nxt ) {
				if ( d[v = graph[i].to] + 1 == l && graph[i].cst < dig ) {
					dig = graph[i].cst;
				}
			}
		}
		if ( dig ) zero = false;
		if ( l == 1 || ! zero ) putchar ( dig ^ '0' );
		for ( int u: curp ) {
			for ( int i = head[u], v; i; i = graph[i].nxt ) {
				if ( d[v = graph[i].to] + 1 == l && graph[i].cst == dig && ! vis[v] ) {
					vis[v] = true, nxtp.push_back ( v ), suf[v] = u;
				}
			}
		}
		curp = nxtp, nxtp.clear ();
	}
	if ( zero ) putchar ( '0' );
	int ans = 1, u;
	for ( u = 1; u ^ n; ++ ans, u = suf[u] );
	printf ( "\n%d\n0", ans ), u = 1;
	do printf ( " %d", ( u = suf[u] ) - 1 ); while ( u ^ n );
	putchar ( '\n' );
	return 0;
}