【分层最短路】Joyride

http://codeforces.com/gym/101873

C

多开一维状态记录时间，d[i][t] = 经过时间t走到节点i的最小花费

每一个状态分别向“原地等待”与“前往下一个节点”转移

代码：

#include <iostream>
#include <algorithm>
#include <queue>
#include <cstring>
#include <vector>
using namespace std;
typedef long long ll;
const int MAX_V = 1005;
const int MAX_E = 2005;
ll p[MAX_V], t[MAX_V];
int N, M, T, pick;
struct Dijkstra {
	struct Edge {
		int to, next;
		ll cost;
	} es[MAX_E];
	struct Node {
		int u;
		ll d, k;
		Node(int u, ll d, ll k) : u(u), d(d), k(k) {}
		bool operator< (const Node& n) const {
			return d > n.d;
		}
	};
	int head[MAX_V];
	int V, E;
	ll d[MAX_V][1005];
	bool vis[MAX_V][1005];
	void init(int V) {
		this->V = V;
		this->E = 0;
		memset(head, -1, sizeof head);
	}
	void addEdge(int u, int v, ll w) {
		es[E].to = v;
		es[E].cost = w;
		es[E].next = head[u];
		head[u] = E++;
	}
	void dijkstra(int s) {
		priority_queue <Node> Q;
		memset(d, 0x3f, sizeof d);
		memset(vis, 0, sizeof(vis));
		d[s][t[1]] = p[1];
		Q.push(Node(s, p[1], t[1]));
		while (!Q.empty()) {
			int u = Q.top().u;
			ll k = Q.top().k;
			Q.pop();
			if (vis[u][k])
				continue;
			vis[u][k] = true;
			if (k + t[u] <= pick && d[u][k + t[u]] > d[u][k] + p[u]) {
				d[u][k + t[u]] = d[u][k] + p[u];
				Q.push(Node(u, d[u][k + t[u]], k + t[u]));
			}
			for (int i = head[u]; i != -1; i = es[i].next) {
				int v = es[i].to;
				ll w = es[i].cost;
				if (k + w + t[v] <= pick && d[v][k + w + t[v]] > d[u][k] + p[v]) {
					d[v][k + w + t[v]] = d[u][k] + p[v];
					Q.push(Node(v, d[v][k + w + t[v]], k + w + t[v]));
				}
			}
		}
	}
} dijk;
int main() {
	scanf("%d%d%d%d", &pick, &N, &M, &T);
	dijk.init(N);
	while (M--) {
		int a, b;
		scanf("%d%d", &a, &b);
		dijk.addEdge(a, b, T);
		dijk.addEdge(b, a, T);
	}
	for (int i = 1; i <= N; i++) {
		scanf("%lld%lld", &t[i], &p[i]);
	}
	dijk.dijkstra(1);
	if (dijk.vis[1][pick])
		printf("%lld\n", dijk.d[1][pick]);
	else
		puts("It is a trap.");
}