UVA 10537

UVA 10537 - The Toll! Revisited

题目链接

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
", 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:
", ++cas);
		printf("%lld
", gao.d[u]);
		gao.print(u);
	}
	return 0;
}

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