UVA - 10537 The Toll! Revisited (最短路变形逆推)

Description

Problem G

Toll! Revisited

Input: Standard Input

Output: Standard Output
Time Limit: 1 Second


Sindbad the Sailor sold 66 silver spoons to the Sultan of Samarkand. The selling was quite easy; but delivering was complicated. The items were transported over land, passing through several towns and villages. Each town and village demanded
an entry toll. There were no tolls for leaving. The toll for entering a village was simply one item. The toll for entering atown was one piece per
20 items carried. For example, to enter a town carrying70 items, you had to pay4 items as toll. The towns and villages were situated strategically between rocks, swamps and rivers, so you could not avoid them.

Figure 1: To reach Samarkand with 66 spoons, traveling through a town followed by two villages, you must start with76 spoons.

Figure 2: The best route to reach X with 39 spoons, starting fromA, is A->b->c->X, shown with arrows in the figure on the left. The best route to reachX with 10 spoons is A->D->X, shown in the figure on the right. The figures display towns as squares and villages as circles.

Predicting the tolls charged in each village or town is quite simple, but finding the best route (the cheapest route) is a real challenge. The best route depends upon the number of items carried. For numbers up to20, villages and towns charge
the same. For large numbers of items, it makes sense to avoid towns and travel through more villages, as illustrated in Figure2.



You must write a program to solve Sindbad’s problem. Given the number of items to be delivered to a certain town or village and a road map, your program must determine the total number of items required at the beginning of the journey that uses a cheapest
route. You will also have to find the cheapest route. If there is more than one such route, print the lexicographically smallest one (A-n-d is smaller thana-n-d).

Input

The input consists of several test cases. Each test case consists of two parts: the roadmap followed by the delivery details.



The first line of the roadmap contains an integer n, which is the number of roads in the map(0 <=
n). Each of the next n lines contains exactly two letters representing the two endpoints of a road. A capital letter represents a town; a lower case letter represents a village. Roads can be traveled in either direction.



Following the roadmap is a single line for the delivery details. This line consists of three things: an integerp (0 < p < 1000000000) for the number of items that must be delivered, a letter for the starting place, and a letter for the place
of delivery. The roadmap is always such that the items can be delivered.



The last test case is followed by a line containing the number -1.

Output

The output consists of three lines for each test case. First line displays the case number, second line shows the number of items required at the beginning of the journey and third line shows the path according to the problem statement above. Actually, the
path contains all the city/village names that Sindbad sees along his journey. Two consecutive city/village names in the path are separated by a hyphen.

Sample Input Output for Sample Input

1 a Z 19 a Z 5 A D D X A b b c c X 39 A X -1

Case 1: 20 a-Z Case 2: 44 A-b-c-X

题意：运送货物需要过路费，进入一个村庄需要缴纳一个单位的货物，而进入一个城镇需要1/20个单位，给你起点和终点，让你求找一条缴纳过路费最少的路线，多条的话要字典序

思路：逆着推最短路，坑的是边界处理，不再是20而是19，没注意，W到死啊，在找字典序的话，ASCII越小就越小

#include<cstdio>
#include<cmath>
#include<ctype.h>
#include<vector>
#include<cstring>
using namespace std;
const int maxn = 55;
typedef long long ll;
const ll INF = 1ll<<62;

vector<int> G[maxn];
bool vis[maxn];
ll dis[maxn];

void dijkstra(int s, int e, ll x, int n) {
	for (int i = 1; i <= n; i++)
		dis[i] = INF;
	dis[s] = x;
	memset(vis, 0, sizeof(vis));
	for (int i = 0; i < n; i++) {
		int index;
		ll Min = INF;
		for (int j = 1; j <= n; j++)
			if (!vis[j] && dis[j] < Min) {
				Min = dis[j];
				index = j;
			}
		if (Min == INF)
			break;
		vis[index] = 1;
		for (int j = 0; j < G[index].size(); j++) {
			ll tmp;
			if (index < 27)
				tmp = (ll) ceil(dis[index]/19.0);
			else tmp = 1;
			if (dis[index]+tmp < dis[G[index][j]] && !vis[G[index][j]])
				dis[G[index][j]] = dis[index]+tmp;
		}
	}
}

int main() {
	int n;
	int u, v;
	int cas = 1;
	ll x;
	char s[10], e[10];
	while (scanf("%d", &n) != EOF && n != -1) {
		for (int i = 1; i <= 52; i++)
			G[i].clear();
		for (int i = 0; i < n; i++) {
			scanf("%s%s", s, e);
			if (isupper(s[0]))
				u = s[0] - 'A' + 1;
			else u = s[0] - 'a' + 27;
			if (isupper(e[0]))
				v = e[0] - 'A' + 1;
			else v = e[0] - 'a' + 27;
			G[u].push_back(v);
			G[v].push_back(u);
		}
		scanf("%lld%s%s", &x, s, e);
		if (isupper(s[0]))
			u = s[0] - 'A' + 1;
		else u = s[0] - 'a' + 27;
		if (isupper(e[0]))
			v = e[0] - 'A' + 1;
		else v = e[0] - 'a' + 27;
		dijkstra(v, u, x, 52);
		printf("Case %d:\n", cas++);
		printf("%lld\n", dis[u]);
		printf("%c", s[0]);
		int cur = u;
		while (cur != v) {
			int Min = 100;
			for (int j = 0; j < G[cur].size(); j++) {
				ll tmp;
				if (G[cur][j] < 27)
					tmp = (ll) ceil(dis[cur]/20.0);
				else tmp = 1;
				if (dis[cur]-tmp == dis[G[cur][j]] && Min > G[cur][j]) 
					Min = G[cur][j];
			}
			cur = Min;
			printf("-%c", cur<27?(cur-1+'A'):(cur-27+'a'));
		}
		printf("\n");
	}
	return 0;
}