GYM 101933D（最短路、二分、dp）

要点

• 非要先来后到暗示多源最短路，求最小的最大值暗示二分
• 二分内部的check是关键，dp处理一下，\(dp[i]\)表示第\(i\)笔订单最早何时送达，如果在ddl之前到不了则\(return\ 0\)。我觉得其中\(time\)变量的维护很好地使复杂度降了一维。
• 第一发WA点：算法看了一遍感觉没有可改的，就把二分的\(r\)调大了，又把\(longlong\)的输入输出改为流，莽试一发就过了……

#include <cstdio>
#include <iostream>
#include <queue>
using namespace std;

typedef long long ll;
const int maxn = 1005, maxm = 5005;
const ll INF = 1e18;

int n, m, k;
ll s[maxn], u[maxn], t[maxn];
struct Edge {
	int to, nxt, cost;
}e[maxm << 1];
int head[maxn], tot;
ll dis[maxn][maxn];

void add(int u, int v, int c) {
	e[++tot].to = v, e[tot].cost = c, e[tot].nxt = head[u], head[u] = tot;
}

void dij(int st) {
	for (int i = 1; i <= n; i++)	dis[st][i] = INF;
	dis[st][st] = 0;
	
	typedef pair<ll, int> pli;
	priority_queue<pli, vector<pli>, greater<pli>> Q;
	Q.push({0, st});

	while (Q.size()) {
		ll d = Q.top().first;
		int p = Q.top().second;
		Q.pop();
		if (d > dis[st][p])	continue;
		for (int i = head[p]; i; i = e[i].nxt) {
			int t = e[i].to;
			if (dis[st][t] > d + e[i].cost) {
				dis[st][t] = d + e[i].cost;
				Q.push({dis[st][t], t});
			}
		}
	}
}

bool ok(ll D) {
	ll dp[maxn];
	dp[0] = 0;
	for (int i = 1; i <= k; i++)
		dp[i] = INF;
	for (int i = 1; i <= k; i++) {
		ll val = dis[1][u[i]];
		ll st = max(t[i], dp[i - 1] + dis[u[i - 1]][1]);
		dp[i] = min(dp[i], st + val);
		if (dp[i] > s[i] + D)	return 0;

		ll time = s[i] + D - st - val;
		if (time < 0)	continue;
		for (int j = i + 1; j <= k; j++) {
			val += dis[u[j - 1]][u[j]];
			if (t[j] > st) {
				time -= t[j] - st;
				st = t[j];
			}
			time = min(time, s[j] + D - st - val);
			if (time < 0)	break;
			dp[j] = min(dp[j], st + val);
		}
	}
	return 1;
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 1, u, v, c; i <= m; i++) {
		scanf("%d %d %d", &u, &v, &c);
		add(u, v, c), add(v, u, c);
	}
	scanf("%d", &k);
	for (int i = 1; i <= k; i++)
		cin >> s[i] >> u[i] >> t[i];

	for (int i = 1; i <= n; i++) {
		dij(i);
	}
	ll l = 0, r = 1e18, ans;
	while (l <= r) {
		ll mid = (l + r) >> 1;
		if (ok(mid))	r = mid - 1, ans = mid;
		else	l = mid + 1;
	}
	cout << ans << '\n';
	return 0;
}