1030 Travel Plan (30分)
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:

4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20
Sample Output:

0 2 3 3 40
只使用dijkstra

#include<iostream>
#include<memory.h>
#include<vector>
#include<algorithm>
using namespace std;
int n, m, st, ed;
const int inf = 0x3f3f3f3f;
const int maxn = 501;//n为顶点数，m为边数，st和ed分别为起点和终点
//g为矩阵，weight为花费
//d记录最短距离，w记录最小花费
int g[maxn][maxn]{inf};
int weight[maxn][maxn];
int w[maxn];
int d[maxn];
bool vist[maxn]{ false };
int path[maxn];//记录父节点
void dijistra(int s)
{fill(w, w+maxn, inf);fill(d, d + maxn, inf);for (int i = 0; i < n; i++)path[i] = i;d[s] = 0;w[s] = 0;for (int i = 0; i < n; i++){int u = -1;int min = inf;for (int j = 0; j < n; j++){if (!vist[j] && d[j] < min){min = d[j];u = j;}}if (u == -1)return;vist[u] = true;for (int j = 0; j < n; j++){if (!vist[j] && g[u][j] != inf){if (d[j] > g[u][j] + d[u]){d[j] = g[u][j] + d[u];w[j] = weight[u][j] + w[u];path[j] = u;}else if (d[j] == g[u][j] + d[u]){if (w[j] > weight[u][j] + w[u]){w[j] = weight[u][j] + w[u];path[j] = u;}}}}}
}
void dfs(int v)
{if (v == st){cout << v << " ";return;}dfs(path[v]);cout << v<<" ";return;
}
int main()
{cin >> n >> m >> st >> ed;fill(g[0], g[0] + maxn * maxn, inf);for (int i = 0; i < m; i++){int u, v;cin >> u >> v;cin >> g[u][v];cin >> weight[u][v];g[v][u] = g[u][v];weight[v][u] = weight[u][v];}dijistra(st);dfs(ed);cout << d[ed]<<" ";cout << w[ed];
}

dijkstra+dfs

#include<iostream>
#include<memory.h>
#include<vector>
using namespace std;
int n, m, st, ed;
const int inf = 0x3f3f3f3f;
const int maxn = 501;//最大顶点数//n为顶点数，m为边数，st和ed分别为起点和终点
//g为矩阵，weight为花费
//d记录最短距离；mincost记录最短路径上的最小花费
int g[maxn][maxn]{inf};
int weight[maxn][maxn];
int mincost = inf;
int d[maxn];
bool vist[maxn]{ false };
vector<int>path[maxn];//前驱
vector<int>temp,path_n;//临时路径和最优路径
void dijistra(int s)
{fill(d, d + maxn, inf);d[s] = 0;for (int i = 0; i < n; i++){int u = -1;int min = inf;for (int j = 0; j < n; j++){if (!vist[j] && d[j] < min){min = d[j];u = j;}}if (u == -1)return;vist[u] = true;for (int j = 0; j < n; j++){if (!vist[j] && g[u][j] != inf){if (d[j] > g[u][j] + d[u]){d[j] = g[u][j] + d[u];path[j].clear();path[j].push_back(u);}else if (d[j] == g[u][j] + d[u]){path[j].push_back(u);	}}}}
}
void dfs(int v)
{if (v == st){temp.push_back(v);int tempcost = 0;for (int i = temp.size() - 1; i > 0; i--){int id = temp[i];int idnext = temp[i - 1];tempcost += weight[id][idnext];}if (mincost > tempcost){mincost = tempcost;path_n = temp;}temp.pop_back();return;}temp.push_back(v);for (int i = 0; i < path[v].size(); i++){dfs(path[v][i]);}temp.pop_back();
}
int main()
{cin >> n >> m >> st >> ed;fill(g[0], g[0] + maxn * maxn, inf);memset(weight, 0, sizeof(weight));for (int i = 0; i < m; i++){int u, v;cin >> u >> v;cin >> g[u][v];cin >> weight[u][v];g[v][u] = g[u][v];weight[v][u] = weight[u][v];}dijistra(st);dfs(ed);for (int i =path_n.size()-1; i >= 0; i--){cout << path_n[i] << " ";}cout << d[ed]<<" "<<mincost;
}