1030 Travel Plan (30 分)

题目链接：

PTA | 程序设计类实验辅助教学平台千名教师建设，万道高质量题目，百万用户拼题的程序设计实验辅助教学平台https://pintia.cn/problem-sets/994805342720868352/problems/994805464397627392

题目描述：

A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 4 positive integers N, M, S, and D, where N (≤500) is the number of cities (and hence the cities are numbered from 0 to N−1); M is the number of highways; S and D are the starting and the destination cities, respectively. Then M lines follow, each provides the information of a highway, in the format:

City1 City2 Distance Cost

where the numbers are all integers no more than 500, and are separated by a space.

Output Specification:

For each test case, print in one line the cities along the shortest path from the starting point to the destination, followed by the total distance and the total cost of the path. The numbers must be separated by a space and there must be no extra space at the end of output.

Sample Input:

Sample Output:

0 2 3 3 40

样例说明：

思路：

1. 最短路径的一道题，要输出最短路径，Dijkstra+DFS， Dijkstra求最短路，DFS打印最短路。

2. 因为同时还要求最小花费，所以如果最短路条数不唯一，则还需要看当前路径是否花费更小。

3. 因为要打印出最短路径，所以要将S到达D点的前序路径都保存下来，再通过DFS深度遍历输出路径。

代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 505, inf = 0x3f3f3f3f;int G[maxn][maxn], d[maxn], c[maxn], cost[maxn][maxn], pre[maxn], minCost; // G邻接矩阵,d最短距离,cost花费,pre保存前序路径
int n, m, S, D;
bool vis[maxn]; // vis当前城市是否访问过void Dijkstra(int s)
{// 初始化d,cfill(d, d + maxn, inf);fill(c, c + maxn, inf);d[s] = 0; // 到自身的距离为0c[s] = 0; // 到自身的花费为0for (int i = 0; i < n; i++) { // 对每个城市都进行一次dijkstraint MIN = inf, u = -1;for (int j = 0; j < n; j++) { // 查找未访问过的城市中距离最小的if (!vis[j] && d[j] < MIN) {MIN = d[j];u = j;}}if (u == -1) // 未找到直接返回return;vis[u] = true; // 标记当前城市已访问for (int j = 0; j < n; j++) { // 遍历n个城市if (!vis[j] && G[u][j] != inf) { // 能直接到达且未访问过的城市if (d[u] + G[u][j] < d[j]) { // 如果当前路径到达的距离比之前的小d[j] = d[u] + G[u][j]; // 更新最短路径c[j] = c[u] + cost[u][j]; // 更新花费pre[j] = u; // 更新前序路径} else if (d[u] + G[u][j] == d[j]) { // 如果最短路不唯一if (c[j] > c[u] + cost[u][j]) { // 如果当前路径花费比之前的小c[j] = c[u] + cost[u][j]; // 更新花费pre[j] = u; // 更新前序路径}}}}}
}void DFS(int u) // 从终点开始递归打印路径
{if (u == S) { // 如果到达起点, 输出起点城市并返回printf("%d ", u);return;}DFS(pre[u]); // 否则递归查找前序路径printf("%d ", u); // 输出经过的城市
}int main()
{fill(G[0], G[0] + maxn * maxn, inf);cin >> n >> m >> S >> D;for (int i = 0; i < m; i++) {int a, b, dis, w;cin >> a >> b >> dis >> w;G[a][b] = G[b][a] = dis; // 距离赋值cost[a][b] = cost[b][a] = w; // 花费赋值}Dijkstra(S);DFS(D); // 打印路径printf("%d %d\n", d[D], c[D]); // 输出最短距离和最小花费return 0;
}

总结：

最短路题目的模板了，同时考虑最短路和最小花费，不管怎么样都先Dijkstra求最短路，然后求最短路的过程中保存最小花费的路径，最后递归打印出来。