uva 10537 - The Toll! Revisited（最短路）

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题目链接:uva 10537 - The Toll! Revisited

从终点向起点做最短路，维护到达节点最少需要多少个。

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 55;const int inf = 0x3f3f3f3f;int IDX(char ch) { if (ch >= 'A' && ch <= 'Z') return ch - 'A'; else return 26 + ch - 'a'; }char RIDX(int x) { if (x < 26) return 'A' + x; else return 'a' + x - 26; }int N = 52, M, S, E, L, V[maxn];ll D[maxn];vector<int> G[maxn], ans;struct State {int u;ll d;State (int u = 0, ll d = 0): u(u), d(d) {}bool operator < (const State& u) const { return d > u.d; }};void init () {for (int i = 0; i <= N; i++) G[i].clear();char s[5], e[5];for (int i = 0; i < M; i++) {scanf("%s%s", s, e);int u = IDX(s[0]), v = IDX(e[0]);G[u].push_back(v);G[v].push_back(u);}scanf("%d%s%s", &L, s, e);S = IDX(s[0]), E = IDX(e[0]);}ll limit (int x, ll w) {if (x >= 26) return w+1;ll l = 0, r = inf;while (l < r) {ll mid = (l + r) >> 1;ll t = mid + w;if (t - (t + 19) / 20 >= w) r = mid;else l = mid + 1;}return w + l;}ll consume(int u, ll d) {if (u >= 26) return 1;else return (d + 19) / 20;}void put (int u) {ans.push_back(u);if (u == E) return;int rec = -1;for (int i = 0; i < G[u].size(); i++) {int v = G[u][i];if (D[u] - consume(v, D[u]) == D[v] && (rec == -1 || rec > v))rec = v;}put(rec);}void dijkstra () {memset(V, 0, sizeof(V));memset(D, inf, sizeof(D));D[E] = L;priority_queue<State> Q;Q.push(State(E, D[E]));while (!Q.empty()) {int u = Q.top().u;Q.pop();if (V[u]) continue;V[u] = 1;ll w = limit(u, D[u]);for (int i = 0; i < G[u].size(); i++) {int v = G[u][i];if (D[v] > w) {D[v] = w;Q.push(State(v, w));}}}printf("%lld\n", D[S]);ans.clear();put(S);for (int i = 0; i < ans.size(); i++) {printf("%c%c", RIDX(ans[i]), i == ans.size()-1 ? '\n' : '-');}}int main () {int cas = 1;while (scanf("%d", &M) == 1 && M != -1) {init ();printf("Case %d:\n", cas++);dijkstra();}return 0;}

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