UVA 11280-flying to Fredericton (shortest way)

Topic Link: UVA 11280-flying to Fredericton

Add one-dimensional means to arrive at the point of docking several times, then Dijsktra do the shortest way.

#include <cstdio> #include <cstring> #include <map> #include <string> #include <vector> #
Include <queue> #include <iostream> #include <algorithm> using namespace std;
typedef pair<int,int> PII;
const int MAXN = 105;

const int inf = 0X3F3F3F3F;
BOOL DONE[MAXN][MAXN];
int N, M, D[MAXN][MAXN];
Vector<pii> G[MAXN];

Map<string, int> T;
	void init () {int D;
	Char CITY1[MAXN], CITY2[MAXN];

	T.clear ();
	scanf ("%d", &n);
		for (int i = 0; i < N; i++) {g[i].clear ();
		scanf ("%s", city1);
	T[city1] = i;
	} scanf ("%d", &m);
		for (int i = 0; i < M; i++) {scanf ("%s%s%d", City1, City2, &d);
		int u = t[city1], v = t[city2];
	G[u].push_back (Make_pair (V, D));
	}}/* bool SPFA (int s, int e) {for (int i = 0; I <= N; i++) for (int j = 0; J <= 101; j + +) d[i][j] = inf;
	memset (inque, 0, sizeof (inque));
	memset (otque, 0, sizeof (otque));

	D[s][0] = 0;
	Queue<pii> Q;
Q.push (Make_pair (s, 0));	while (!
		Q.empty ()) {int u = Q.front (). First;
		int t = Q.front (). Second;

		Q.pop ();
		Inque[u][t] = 0;
		otque[u][t]++;
		if (Otque[u][t] > N) return false;
			for (int i = 0; i < g[u].size (); i++) {int v = g[u][i].first, w = g[u][i].second;

			if (d[v][t+1] > D[u][t] + w) d[v][t+1] = d[u][t] + W;
				if (t <= &&!inque[v][t+1]) {inque[v][t+1] = 1;
			Q.push (Make_pair (V, T + 1));
}}} return true;
	} */struct state {int u, T, D;  state (int u = 0, int t = 0, int d = 0): U (U), T (t), D (d) {} BOOL operator < (const state& a) const {return d > A.D;

}
};
	void Dijkstra (int s) {for (int i = 0; i < N; i++) for (int j = 0; J <=; j + +) d[i][j] = inf;
	D[s][0] = 0;

	memset (done, 0, sizeof (done));
	Priority_queue<state> Q;

	Q.push (state (S, 0, d[s][0])); while (!
		Q.empty ()) {int u = q.top (). u;
		int t = q.top (). T;

		Q.pop ();
		if (done[u][t]) continue;

		Done[u][t] = true; for (int i = 0; i < g[u].size ();
			i++) {int v = g[u][i].first, w = g[u][i].second;
				if (d[v][t+1] > D[u][t] + W) {d[v][t+1] = D[u][t] + W;
			Q.push (State (V, t+1, d[v][t+1]));
	}}}} int main () {int cas;
	scanf ("%d", &cas);
		for (int kcas = 1; kcas <= cas; kcas++) {if (Kcas! = 1) printf ("\ n");
		Init ();

		Dijkstra (0);
		printf ("Scenario #%d\n", Kcas);
		int q, x;
		scanf ("%d", &q);
			while (q--) {scanf ("%d", &x);
			int ans = inf;
			x = min (x+1, N-1);
			for (int i = 0; I <= x; i++) ans = min (ans, d[n-1][i]);
			if (ans! = inf) printf ("Total cost of flight (s) is $%d\n", ans);
		else printf ("No satisfactory flights\n");
}} return 0;
 }