Luogu1261: 服务器储存信息问题

题面

传送门

Sol

我们可以考虑每种\(rank\)的点\(u\)会被哪些点\(v\)感兴趣
如果\(dis[u][v]<\)所有满足\(rank\)大于\(rank[u]\)的点到\(v\)的距离
那么\(v\)一定对\(u\)感兴趣

我们可以先\(10\)遍最短路，处理出每种\(rank\)的点到所有点的最短路
计\(f[i][j]\)表示满足\(i\le rank\)的点到所有点的最短路

然后每次从一个点\(u\)出发，寻找对它感兴趣的点\(v\)
如果发现\(f[rank[u]+1][v]\le dis[u][v]\)
那么\(v\)不对\(u\)感兴趣，而且所有\(v\)能松弛到的点也都不会
那么可以\(SPFA\)，每次不满足要求了就不丢进队列

复杂度是\(O(ans+10n)\)的，可以接受

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
const int _(3e4 + 5);
const int INF(1e9);
typedef long long ll;

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, m, rk[_], first[_], cnt, f[15][_], vis[_], ans, dis[_], use[_];
struct Edge{
	int to, next, w;
} edge[_ * 10];
struct Data{
	int u, d;

	IL int operator <(RG Data A) const{
		return d > A.d;
	}
};
priority_queue <Data> Q;
queue <int> Deq;

IL void Add(RG int u, RG int v, RG int w){
	edge[cnt] = (Edge){v, first[u], w}, first[u] = cnt++;
}

IL void Dijkstra(RG int p, RG int *dis){
	for(RG int i = 1; i <= n; vis[i] = 0, ++i)
		if(rk[i] == p) dis[i] = 0, Q.push((Data){i, 0});
		else dis[i] = INF;
	while(!Q.empty()){
		RG Data x = Q.top(); Q.pop();
		if(vis[x.u]) continue;
		vis[x.u] = 1;
		for(RG int e = first[x.u]; e != -1; e = edge[e].next){
			RG int v = edge[e].to, w = edge[e].w;
			if(x.d + w < dis[v]) Q.push((Data){v, dis[v] = x.d + w});
		}
	}
}

IL int SPFA(RG int x){
	RG int ret = 0;
	for(RG int i = 1; i <= n; ++i) dis[i] = INF, use[i] = vis[i] = 0;
	Deq.push(x), dis[x] = 0, vis[x] = 1;
	while(!Deq.empty()){
		RG int u = Deq.front(); Deq.pop();
		if(!use[u]) use[u] = 1, ++ret;
		for(RG int e = first[u]; e != -1; e = edge[e].next){
			RG int v = edge[e].to, w = edge[e].w;
			if(dis[u] + w < dis[v]){
				dis[v] = dis[u] + w;
				if(!vis[v] && f[rk[x] + 1][v] > dis[v]) vis[v] = 1, Deq.push(v);
			}
		}
		vis[u] = 0;
	}
	return ret;
}

int main(RG int argc, RG char* argv[]){
	n = Input(), m = Input();
	for(RG int i = 1; i <= n; ++i) rk[i] = Input(), first[i] = -1, f[11][i] = INF;
	for(RG int i = 1; i <= m; ++i){
		RG int u = Input(), v = Input(), w = Input();
		Add(u, v, w), Add(v, u, w);
	}
	for(RG int i = 10; i; --i){
		Dijkstra(i, f[i]);
		for(RG int j = 1; j <= n; ++j) f[i][j] = min(f[i][j], f[i + 1][j]);
	}
	for(RG int i = 1; i <= n; ++i) ans += SPFA(i);
	printf("%d\n", ans);
    return 0;
}