Bzoj4016/洛谷P2993 [FJOI2014] 最短路径树问题（最短路径问题+长链剖分/点分治）

题面

Bzoj

洛谷

题解

首先把最短路径树建出来（用$Dijkstra$，没试过$SPFA$$\leftarrow$它死了），然后问题就变成了一个关于深度的问题，可以用长链剖分做，所以我们用点分治来做（滑稽）。

有一点要说，这一题数据比较水，如果不用字典序的话，也可以过。如何建立字典序呢？其实我们从$1$号节点开始遍历路径树（不是最短路径树），令一个点的第一关键字是点权，如果点权相等就按照编号大小为第二关键字，维护一个二元组就好了。

点分治时记两个数组$S[i]$和$num[i]$，表示经过$i$个点的路径最大是多少以及在这个情况下有多少条路径。

之前找重心调了好久。

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
using std::pair; using std::sort;
using std::priority_queue;
using std::vector; using std::greater;
typedef long long ll;
typedef pair<int, int> pii;
template<typename T>
void read(T &x) {
    int flag = 1; x = 0; char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') flag = -flag; ch = getchar(); }
    while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); x *= flag;
}
const int N = 3e4 + 10, Inf = 1 << 30;
int n, m, k, from[N], dist[N], MX, tot;
int cnt, to[N << 1], nxt[N << 1], dis[N << 1];
bool vis[N];
vector<pii> G[N]; priority_queue< pii, vector<pii>, greater<pii> > q;
void addEdge(int u, int v, int w) {
	to[++cnt] = v, nxt[cnt] = from[u], dis[cnt] = w, from[u] = cnt;
}
void dijk(int s) {
	memset(dist, 0x7777777f, sizeof dist);
	dist[s] = 0, q.push((pii){0, s});
	while(q.size()) {
		int u = q.top().second; q.pop();
		if(vis[u]) continue; vis[u] = true;
		for(int i = 0; i < G[u].size(); ++i) {
			int v = G[u][i].first, w = G[u][i].second + dist[u];
			if(dist[v] > w) dist[v] = w, q.push((pii){dist[v], v});
		}
	}
}
void init(int u) {
	vis[u] = 1;
	for(int i = 0; i < G[u].size(); ++i) {
		int v = G[u][i].first, w = G[u][i].second;
		if(vis[v] || w + dist[u] != dist[v]) continue;
		addEdge(u, v, w), addEdge(v, u, w), init(v);
	}
}
int Size, tmp, p, siz[N], maxnow, S[N], num[N];
inline void upt(int &a, int b) { if(a < b) a = b; }
void getrt(int u, int f) {
	int max_part = 0; siz[u] = 1;
	for(int i = from[u]; i; i = nxt[i]) {
		int v = to[i]; if(vis[v] || v == f) continue;
		getrt(v, u); siz[u] += siz[v];
		upt(max_part, siz[v]);
	} upt(max_part, Size - siz[u]);
	if(max_part < tmp) p = u, tmp = max_part;
}
void calc(int u, int f, int now) {
	upt(maxnow, now);
	if(now == k - 1) {
		if(dist[u] == MX) ++tot;
		else if(dist[u] > MX) MX = dist[u], tot = 1;
		return ;
	}
	int nowans = -1;
	if(S[k - 1 - now] != -1) nowans = dist[u] + S[k - 1 - now];
	if(nowans == MX) tot += num[k - 1 - now];
	else if(nowans > MX) MX = nowans, tot = num[k - 1 - now];
	for(int i = from[u]; i; i = nxt[i]) {
		int v = to[i]; if(vis[v] || v == f) continue;
		dist[v] = dist[u] + dis[i], calc(v, u, now + 1);
	}
}
void update(int u, int f, int now) {
	if(now == k - 1) return ;
	if(S[now] == dist[u]) ++num[now];
	else upt(S[now], dist[u]), num[now] = 1;
	for(int i = from[u]; i; i = nxt[i]) {
		int v = to[i]; if(vis[v] || v == f) continue;
		update(v, u, now + 1);
	}
}
void doit(int x) {
	p = 0, tmp = Inf, getrt(x, 0), vis[p] = 1, maxnow = 0;
	for(int i = from[p]; i; i = nxt[i]) {
		int v = to[i]; if(vis[v]) continue;
		dist[v] = dis[i], calc(v, p, 1), update(v, p, 1);
	}
	for(int i = 1; i <= maxnow; ++i) S[i] = -1, num[i] = 0;
	for(int i = from[p]; i; i = nxt[i]) {
		int v = to[i]; if(vis[v]) continue;
		Size = siz[v], doit(v);
	}
}
int main () {
	read(n), read(m), read(k);
	for(int i = 1, u, v, w; i <= m; ++i) {
		read(u), read(v), read(w);
		G[u].push_back((pii){v, w});
		G[v].push_back((pii){u, w});
	}
	for(int i = 1; i <= n; ++i) sort(G[i].begin(), G[i].end());
	dijk(1), memset(vis, 0, sizeof vis), init(1);
	Size = n, memset(vis, 0, sizeof vis);
	memset(dist, 0, sizeof dist), doit(1);
	printf("%d %d\n", MX, tot);
	return 0;
}