美团2018年CodeM大赛-初赛B轮 B 配送（最短路）

美团2018年CodeM大赛-初赛B轮 B 配送

题意

题解

对于每个任务，只要从上个任务的终点出发即可。
时间、地点很少，可以算出每个地点-时间的最小花费。
以题目描述的起点终点起始结束时间建图，很暴力的跑最短路即可。

代码

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define rep(i, a, b) for(int i=(a); i<(b); i++)
#define sz(a) (int)a.size()
#define de(a) cout << #a << " = " << a << endl
#define dd(a) cout << #a << " = " << a << " "
#define all(a) a.begin(), a.end()
#define endl "\n"
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
//---

const int inf = 2e9;

int n, m, k;
struct Task {
	int pos;
	string tim;
	bool operator < (const Task &c) const {
		return tim < c.tim;
	}
}task[11];
struct Edge {
	int v, p;
	string st, ed;
	Edge(int v, int p, string st, string ed) : v(v), p(p), st(st), ed(ed) {}
};
vector<Edge> g[111];
map<string, int> dis[111];

void dij(int pos, string tim) {
	rep(i, 1, k+1) dis[i].clear();
	dis[pos][tim] = 0;
	priority_queue<pair<int, pair<int, string> > > que;
	que.push(mp(0, mp(pos, tim)));
	while(!que.empty()) {
		auto u = que.top(); que.pop();
		int d = -u.fi;
		pos = u.se.fi; tim = u.se.se;
		if(d != dis[pos][tim]) continue;
		for(auto i : g[pos]) {
			int v = i.v, p = i.p;
			int t = dis[pos][tim] + p;
			if(tim < i.st && (!dis[v].count(i.ed) || dis[v][i.ed] > t)) {
				dis[v][i.ed] = t;
				que.push(mp(-t, mp(v, i.ed)));
			}
		}
	}
}

int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(0);
	cin >> n >> m >> k;
	rep(i, 0, n) {
		string s1, s2;
		cin >> task[i].pos >> s1 >> s2;
		task[i].tim = s1 + " " + s2;
	}
	sort(task, task+n);
	rep(i, 0, m) {
		int u, v, p;
		string st, ed;
		cin >> u >> v >> p >> st >> ed;
		rep(j, 1, 8) {
			string t1 = "2018.07.0" + to_string(j) + " " + st;
			string t2 = "2018.07.0" + to_string(j) + " " + ed;
			g[u].pb(Edge(v, p, t1, t2));
		}
	}
	int pos = 1, ans = 0;
	string tim = "2018.06.30 23:59:59.999";
	rep(i, 0, n) {
		dij(pos, tim);
		int res = inf;
		for(auto j : dis[task[i].pos]) {
			if(j.fi < task[i].tim) res = min(res, j.se);
			else break;
		}
		if(res == inf) {
			ans = inf;
			break;
		}
		ans += res;
		pos = task[i].pos;
		tim = task[i].tim;
	}
	if(ans == inf) ans = -1;
	cout << ans << endl;
	return 0;
}