题意：给定一个图，你要从 s 到达 t，当经过大写字母时，要交 ceil(x /20)的税，如果经过小写字母，那么交 1的税，问你到达 t 后还剩下 c 的，那么最少要带多少，并输出一个解，如果多个解，则输出字典序最小的。
析：最短路，逆推，d[i] 表示的是从 i 到时 t 最少要带多少，然后就能顺利的推出从 s 开始时要带多少，然后打印路径，每次取最小的字母即可。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 10;
const int mod = 1000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r > 0 && r <= n && c > 0 && c <= m;
} struct Edge{
int from, to;
LL dist;
};
struct HeapNode{
LL d; int u;
bool operator < (const HeapNode &p) const{
return d > p.d;
}
}; int ID(char ch){
if(islower(ch)) return ch - 'a' + 26;
return ch - 'A';
} char invID(int x){ return x < 26 ? 'A' + x: 'a' + x - 26; } struct Dijkstra{
int n, m;
vector<Edge> edges;
vector<int> G[maxn];
bool done[maxn];
LL d[maxn]; void init(int n){
this->n = n;
for(int i = 0; i < n; ++i) G[i].cl;
edges.cl;
} void addEdge(int u, int v, LL d){
edges.pb((Edge){u, v, d});
m = edges.sz;
G[u].pb(m-1);
} void dijkstra(int s, LL val){
priority_queue<HeapNode> pq;
fill(d, d + n, LNF);
d[s] = val; ms(done, 0);
pq.push((HeapNode){0, s});
while(!pq.empty()){
HeapNode x = pq.top(); pq.pop();
int u = x.u;
if(done[u]) continue;
done[u] = 1;
for(int i = 0; i < G[u].sz; ++i){
Edge &e = edges[G[u][i]];
if(u > 25 && d[e.to] >= d[u] + 1){
e.dist = edges[G[u][i]^1].dist = 1LL;
d[e.to] = d[u] + 1;
pq.push((HeapNode){d[e.to], e.to});
}
else if(u < 26){
LL l = d[u], r = d[u] * 20;
while (l < r) {
LL m = (l + r) / 2;
LL x = m - ceil(m / 20.);
if (x < d[u]) l = m + 1;
else r = m;
}
if(d[e.to] >= l){
e.dist = edges[G[u][i]^1].dist = l - d[u];
d[e.to] = l;
pq.push((HeapNode){d[e.to], e.to});
}
}
}
}
} void solve(int m, int s, int t){
dijkstra(t, m);
printf("%lld\n", d[s]);
while(s != t){
printf("%c-", invID(s));
int mmin = INF;
for(int i = 0; i < G[s].sz; ++i){
Edge &e = edges[G[s][i]];
if(d[s] == d[e.to] + e.dist) mmin = min(mmin, e.to);
}
s = mmin;
}
printf("%c\n", invID(s));
}
}; Dijkstra dij; int main(){
int kase = 0;
while(scanf("%d", &n) == 1 && n != -1){
char s1[5], s2[5];
dij.init(60);
for(int i = 0; i < n; ++i){
scanf("%s %s", s1, s2);
dij.addEdge(ID(s1[0]), ID(s2[0]), 0LL);
dij.addEdge(ID(s2[0]), ID(s1[0]), 0LL);
}
scanf("%d %s %s", &n, s1, s2);
printf("Case %d:\n", ++kase);
dij.solve(n, ID(s1[0]), ID(s2[0]));
}
return 0;
}