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UVA 10537 - The Toll! Revisited(dijstra扩展)
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,如果有多种方案,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需要这么多货物,然后反向建图,从终点向起点反向做一遍
这题被坑了。。并不是输出的城市才存在,比如下面这组样例
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\n", 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:\n", ++cas); printf("%lld\n", gao.d[u]); gao.print(u); } return 0; }
UVA 10537 - The Toll! Revisited(dijstra扩展)
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