UVa12661 Funny Car Racing (最短路)

Posted 2020-08-10

链接：http://vjudge.net/problem/UVA-12661

 

分析：带权图最短路。仍然调用标准的Dijkstra算法，只是在松弛操作时需要分情况讨论。

 

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <vector>
 4 #include <queue>
 5 using namespace std;
 6 
 7 const int maxn = 300 + 5;
 8 
 9 struct Edge {
10     int from, to, open, close, needTime;
11     Edge(int from, int to, int open, int close, int needTime):
12         from(from), to(to), open(open), close(close), needTime(needTime) {}
13 };
14 
15 struct HeapNode {
16     int d, u;
17     bool operator < (const HeapNode& rhs) const {
18         return d > rhs.d;
19     }
20 };
21 
22 struct Dijkstra {
23     int n, m;
24     vector<Edge> edges;
25     vector<int> G[maxn];
26     bool vis[maxn];
27     int d[maxn];
28 
29     void init(int n) {
30         this -> n = n;
31         for (int i = 0; i < n; i++) G[i].clear();
32         edges.clear();
33     }
34 
35     void AddEdge(int from, int to, int open, int close, int needTime) {
36         edges.push_back(Edge(from, to, open, close, needTime));
37         m = edges.size();
38         G[from].push_back(m - 1);
39     }
40 
41     bool check(int ut, int& vt, Edge e) {
42         if (e.needTime > e.open) return false;
43         int T = e.open + e.close;
44         if (ut % T + e.needTime > e.open)
45             ut += T - (ut % T);
46         if (ut + e.needTime < vt) {
47             vt = ut + e.needTime;
48             return true;
49         }
50         return false;
51     }
52 
53     int dijkstra(int s, int t) {
54         memset(vis, 0, sizeof(vis));
55         memset(d, 0x3f, sizeof(d));
56         priority_queue<HeapNode> q;
57         d[s] = 0;
58         q.push((HeapNode){0, s});
59         while (!q.empty()) {
60             int u = q.top().u; q.pop();
61             if (vis[u]) continue; vis[u] = 1;
62             if (u == t) break;
63             for (int i = 0; i < G[u].size(); i++) {
64                 Edge& e = edges[G[u][i]];
65                 if (check(d[u], d[e.to], e)) q.push((HeapNode){d[e.to], e.to});
66             }
67         }
68         return d[t];
69     }
70 };
71 
72 Dijkstra dij;
73 int n, m, s, t;
74 
75 int main() {
76     int kase = 0;
77     while (scanf("%d%d%d%d", &n, &m, &s, &t) == 4) {
78         dij.init(n);
79         int u, v, a, b, t2;
80         while (m--) {
81             scanf("%d%d%d%d%d", &u, &v, &a, &b, &t2);
82             dij.AddEdge(u - 1, v - 1, a, b, t2);
83         }
84         printf("Case %d: %d\n", ++kase, dij.dijkstra(s - 1, t - 1));
85     }
86     return 0;
87 }