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 ", 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 ", 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: ", ++kase); dij.solve(n, ID(s1[0]), ID(s2[0])); } return 0; }