UVA 11280 - Flying to Fredericton（最短路）

UVA 11280 - Flying to Fredericton

题目链接

题意：给定一些国家，和两个国家间的花费，现在有一些询问，询问每次最多转k次飞机，最小花费

思路：dijkstra变形，多开一维表示转机次数即可

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <iostream>
#include <string>
#include <map>
using namespace std;

const int MAXNODE = 505;
const int MAXEDGE = 5005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge {
	int u, v;
	Type dist;
	Edge() {}
	Edge(int u, int v, Type dist) {
		this->u = u;
		this->v = v;
		this->dist = dist;
	}
};

struct HeapNode {
	Type d;
	int ti;
	int u;
	HeapNode() {}
	HeapNode(Type d, int u, int ti) {
		this->d = d;
		this->ti = ti;
		this->u = u;
	}
	bool operator < (const HeapNode& c) const {
		return d > c.d;
	}
};

struct Dijkstra {
	int n, m;
	Edge edges[MAXEDGE];
	int first[MAXNODE];
	int next[MAXEDGE];
	bool done[MAXNODE][MAXNODE];
	Type d[MAXNODE][MAXNODE];
	int p[MAXNODE];

	void init(int n) {
		this->n = n;
		memset(first, -1, sizeof(first));
		m = 0;
	}

	void add_Edge(int u, int v, Type dist) {
		edges[m] = Edge(u, v, dist);
		next[m] = first[u];
		first[u] = m++;
	}

	void dijkstra(int s) {
		priority_queue<HeapNode> Q;
		for (int i = 0; i < n; i++) 
			for (int j = 0; j <= 100; j++)
				d[i][j] = INF;
		d[s][0] = 0;
		memset(done, false, sizeof(done));
		Q.push(HeapNode(0, s, 0));
		while (!Q.empty()) {
			HeapNode x = Q.top(); Q.pop();
			int u = x.u;
			int ti = x.ti;
			if (done[u][ti]) continue;
			done[u][ti] = true;
			for (int i = first[u]; i != -1; i = next[i]) {
				Edge& e = edges[i];
				if (d[e.v][ti + 1] > d[u][ti] + e.dist) {
					d[e.v][ti + 1] = d[u][ti] + e.dist;
					Q.push(HeapNode(d[e.v][ti + 1], e.v, ti + 1));
				}
			}
		}
	}
} gao;

int t, n;
map<string, int> hash;
string name;

int main() {
	int cas = 0;
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		gao.init(n);
		hash.clear();
		for (int i = 0; i < n; i++) {
			cin >> name;
			hash[name] = i;
		}
		int m;
		scanf("%d", &m);
		string u, v;
		int d;
		while (m--) {
			cin >> u >> v >> d;
			gao.add_Edge(hash[u], hash[v], d);
		}
		gao.dijkstra(0);
		printf("Scenario #%d\n", ++cas);
		int q, num;
		scanf("%d", &q);
		while (q--) {
			scanf("%d", &num);
			num++;
			if (num > n - 1) num = n - 1;
			int ans = INF;
			for (int i = 0; i <= num; i++)
				ans = min(ans, gao.d[n - 1][i]);
			if (ans == INF) printf("No satisfactory flights\n");
			else printf("Total cost of flight(s) is $%d\n", ans);
		}
		if (t) printf("\n");
	}
	return 0;
}