旅游规划
作者：互联网
旅游规划
有了一张自驾旅游路线图，你会知道城市间的高速公路长度、以及该公路要收取的过路费。现在需要你写一个程序，帮助前来咨询的游客找一条出发地和目的地之间的最短路径。如果有若干条路径都是最短的，那么需要输出最便宜的一条路径。
输入格式:
输入说明：输入数据的第1行给出4个正整数N、M、S、D，其中N（2 ≤ N ≤ 500）是城市的个数，顺便假设城市的编号为0~(N−1)；M是高速公路的条数；S是出发地的城市编号；D是目的地的城市编号。随后的M行中，每行给出一条高速公路的信息，分别是：城市1、城市2、高速公路长度、收费额，中间用空格分开，数字均为整数且不超过500。输入保证解的存在。
输出格式:
在一行里输出路径的长度和收费总额，数字间以空格分隔，输出结尾不能有多余空格。
输入样例:
4 5 0 3 0 1 1 20 1 3 2 30 0 3 4 10 0 2 2 20 2 3 1 20
输出样例:
3 40
解题思路
这道题其实就是在Dijkstra算法的基础上进行改进。在求最短路径的基础上，再加上一个求最少费用，当然最小路径的优先级比最小费用要高，所有只有在两条路径长度相同时，才判断这两条路径哪一条所需费用最小，然后选择费用最小的那一条路径。
下面给出用邻接表的存图方式，以及最原始的Dijkstra算法的AC代码：
1 #include <cstdio>
2 #include <iostream>
3 #include <algorithm>
4
5 const int INF = 0x3f3f3f3f;
6
7 struct AdjVNode {
8 int adjVer;
9 int len, cost;
10 AdjVNode *next;
11 };
12
13 struct LGraph {
14 AdjVNode **G;
15 int verN, edgeN;
16 };
17
18 LGraph *createLGraph(int n, int m);
19 void Dijkstra(LGraph *graph, int src, int dest);
20
21 int main() {
22 int n, m, src, dest;
23 scanf("%d %d %d %d", &n, &m, &src, &dest);
24 LGraph *graph = createLGraph(n, m);
25 Dijkstra(graph, src, dest);
26
27 return 0;
28 }
29
30 LGraph *createLGraph(int n, int m) {
31 LGraph *graph = new LGraph;
32 graph->verN = n;
33 graph->edgeN = m;
34 graph->G = new AdjVNode*[graph->verN];
35 for (int i = 0; i < graph->verN; i++) {
36 graph->G[i] = NULL;
37 }
38
39 for (int i = 0; i < graph->edgeN; i++) {
40 int v, w, len, cost;
41 scanf("%d %d %d %d", &v, &w, &len, &cost);
42 AdjVNode *tmp = new AdjVNode;
43 tmp->adjVer = w;
44 tmp->len = len;
45 tmp->cost = cost;
46 tmp->next = graph->G[v];
47 graph->G[v] = tmp;
48
49 tmp = new AdjVNode;
50 tmp->adjVer = v;
51 tmp->len = len;
52 tmp->cost = cost;
53 tmp->next = graph->G[w];
54 graph->G[w] = tmp;
55 }
56
57 return graph;
58 }
59
60 void Dijkstra(LGraph *graph, int src, int dest) {
61 int dist[graph->verN], cost[graph->verN];
62 std::fill(dist, dist + graph->verN, INF);
63 std::fill(cost, cost + graph->verN, 0);
64 bool vis[graph->verN] = {false};
65
66 vis[src] = true;
67 dist[src] = 0;
68 for (AdjVNode *w = graph->G[src]; w; w = w->next) {
69 dist[w->adjVer] = w->len;
70 cost[w->adjVer] = w->cost;
71 }
72
73 while (true) {
74 int minV, min = INF;
75 for (int v = 0; v < graph->verN; v++) {
76 if (!vis[v] && dist[v] < min) {
77 min = dist[v];
78 minV = v;
79 }
80 }
81
82 vis[minV] = true;
83 if (vis[dest]) break;
84 for (AdjVNode *w = graph->G[minV]; w; w = w->next) {
85 if (dist[w->adjVer] > dist[minV] + w->len) {
86 dist[w->adjVer] = dist[minV] + w->len;
87 cost[w->adjVer] = cost[minV] + w->cost;
88 }
89 else if (dist[w->adjVer] == dist[minV] + w->len) {
90 cost[w->adjVer] = std::min(cost[w->adjVer], cost[minV] + w->cost);
91 }
92 }
93 }
94
95 printf("%d %d\n", dist[dest], cost[dest]);
96 }
再给出我平时用std::vector来存图，同时用std::priority_queue改进后的Dijkstra算法的AC代码：
1 #include <cstdio>
2 #include <iostream>
3 #include <map>
4 #include <queue>
5 #include <vector>
6 #include <algorithm>
7
8 const int INF = 0x3f3f3f3f;
9
10 typedef std::pair<int, int> Pair;
11
12 struct Edge {
13 int to, len, cost;
14 };
15
16 struct Graph {
17 std::vector<std::vector<Edge> > G;
18 int verN, edgeN;
19 };
20
21 void Dijkstra(Graph *graph, int src, int dest);
22
23 int main() {
24 int n, m, src, dest;
25 scanf("%d %d %d %d", &n, &m, &src, &dest);
26 Graph *graph = new Graph;
27 graph->verN = n;
28 graph->edgeN = m;
29 graph->G.resize(graph->verN);
30 for (int i = 0; i < graph->edgeN; i++) {
31 int v, w, len, cost;
32 scanf("%d %d %d %d", &v, &w, &len, &cost);
33 graph->G[v].push_back({w, len, cost});
34 graph->G[w].push_back({v, len, cost});
35 }
36
37 Dijkstra(graph, src, dest);
38
39 return 0;
40 }
41
42 void Dijkstra(Graph *graph, int src, int dest) {
43 int dist[graph->verN], cost[graph->verN];
44 std::fill(dist, dist + graph->verN, INF);
45 std::fill(cost, cost + graph->verN, INF);
46 bool vis[graph->verN] = {false};
47 std::priority_queue<Pair, std::vector<Pair>, std::greater<Pair> > pq;
48 dist[src] = cost[src] = 0;
49 pq.push(Pair(0, src));
50
51 while (!pq.empty()) {
52 int v = pq.top().second;
53 pq.pop();
54 if (vis[v]) continue;
55 vis[v] = true;
56 if (vis[dest]) break;
57
58 for (std::vector<Edge>::iterator it = graph->G[v].begin(); it != graph->G[v].end(); it++) {
59 if (!vis[it->to] && dist[v] + it->len < dist[it->to]) {
60 dist[it->to] = dist[v] + it->len;
61 cost[it->to] = cost[v] + it->cost;
62 pq.push(Pair(dist[it->to], it->to));
63 }
64 else if (!vis[it->to] && dist[v] + it->len == dist[it->to]) {
65 cost[it->to] = std::min(cost[v] + it->cost, cost[it->to]);
66 }
67 }
68 }
69
70 printf("%d %d", dist[dest], cost[dest]);
71 }
标签：dist,int,graph,len,cost,旅游,规划,verN 来源： https://www.cnblogs.com/onlyblues/p/14819725.html