UVA 10537 - The Toll! Revisited（dijstra扩张）

UVA 10537 - The Toll! Revisited

题意：给定一个无向图，大写字母是城市，小写字母是村庄，经过城市交过路费为当前货物的%5，路过村庄固定交1，给定起点终点和到目标地点要剩下的货物，问最少要带多少货物上路。并输出路径，假设有多种方案。要求字典序最小

思路：dijstra的逆向运用。d数组含义变成到该结点至少须要这么多货物，然后反向建图，从终点向起点反向做一遍

这题被坑了。。并非输出的城市才存在。比方以下这组例子

0

1 A A

应该输出

1

A

代码：

#include <cstdio>

#include <cstring>

#include <vector>

#include <queue>

#include <cmath>

using namespace std;

const int MAXNODE = 105;

typedef long long Type;

const Type INF = (1LL<<61);

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 u;

	HeapNode() {}

	HeapNode(Type d, int u) {

		this->d = d;

		this->u = u;

	}

	bool operator < (const HeapNode& c) const {

		return d > c.d;

	}

};

char to[255];

struct Dijkstra {

	int n, m;

	vector<Edge> edges;

	vector<int> g[MAXNODE];

	bool done[MAXNODE];

	Type d[MAXNODE];

	int p[MAXNODE];

	void init(int tot) {

		n = tot;

		for (int i = 0; i < n; i++)

			g[i].clear();

		edges.clear();

	}

	void add_Edge(int u, int v, Type dist) {

		edges.push_back(Edge(u, v, dist));

		m = edges.size();

		g[u].push_back(m - 1);

	}

	void print(int e) {

		if (p[e] == -1) {

			printf("%c\n", to[e]);

			return;

		}

		printf("%c-", to[e]);

		print(edges[p[e]].u);

	}

	void dijkstra(Type start, int s) {

		priority_queue<HeapNode> Q;

		for (int i = 0; i < n; i++) d[i] = INF;

		d[s] = start;

		p[s] = -1;

		memset(done, false, sizeof(done));

		Q.push(HeapNode(start, s));

		while (!Q.empty()) {

			HeapNode x = Q.top(); Q.pop();

			int u = x.u;

			if (done[u]) continue;

			done[u] = true;

			for (int i = 0; i < g[u].size(); i++) {

				Edge& e = edges[g[u][i]];

				Type need;

				if (e.dist) need = (Type)ceil(d[u] * 1.0 / 19 * 20);

				else need = d[u] + 1;

				if (d[e.v] > need || (d[e.v] == need && to[u] < to[edges[p[e.v]].u])) {

					d[e.v] = need;

					p[e.v] = g[u][i];

					Q.push(HeapNode(d[e.v], e.v));

				}

			}

		}

	}

} gao;

typedef long long ll;

int n, m, vis[255];

int main() {

	int cas = 0;

	for (int i = 0; i < 26; i++) {

		vis['A' + i] = i;

		to[i] = 'A' + i;

	}

	for (int i = 0; i < 26; i++) {

		vis['a' + i] = i + 26;

		to[i + 26] = 'a' + i;

	}

	while (~scanf("%d", &m) && m != -1) {

		gao.init(52);

		char a[2], b[2];

		int u, v;

		while (m--) {

			scanf("%s%s", a, b);

			u = vis[a[0]], v = vis[b[0]];

			gao.add_Edge(u, v, a[0] < 'a');

			gao.add_Edge(v, u, b[0] < 'a');

		}

		ll need;

		scanf("%lld%s%s", &need, a, b);

		u = vis[a[0]]; v = vis[b[0]];

		gao.dijkstra(need, v);

		printf("Case %d:\n", ++cas);

		printf("%lld\n", gao.d[u]);

		gao.print(u);

	}

	return 0;

}

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