1.题目要求最短路径，最短路径不唯一时，求花费最小的路径。
2.仍然是先使用Dijkstra求出最短距离，然后使用深度搜索遍历，以最短距离作为条件进行剪枝，最后求出最短距离的基础上耗费最小的路径。
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
AC代码：
[c language=”++”]
//#include<string>
//#include <iomanip>
#include<vector>
#include <algorithm>
//#include<stack>
#include<set>
#include<queue>
#include<map>
//#include<unordered_set>
//#include<unordered_map>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
using namespace std;
/*
4 5 0 3
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20
4 5 0 0
0 1 1 20
1 3 2 30
0 3 4 10
0 2 2 20
2 3 1 20
*/
struct neighbor
{
int idx;
int w;
long long dist;
neighbor() :idx(-1), w(-1), dist(-1){};
neighbor(int a, int b, int c) :idx(a), w(b), dist(c){};
};
struct vertexNode{
vector<neighbor> nb;
bool visited;
bool sured;
long long dist;
vertexNode(): nb(0), visited(false), sured(false), dist(INT_MAX){};
};
void dfs(int now,int t,int dist,int minDist,int cost,vector<int>& path,vector<pair<int,vector<int>>>& ans,vector<vertexNode>& v,vector<bool>&used)
{
if (dist > minDist) return;
if (now == t && dist == minDist)
{
ans.push_back({ cost, path });
}
else
{
for (int i = 0; i < v[now].nb.size(); i++)
{
int q = v[now].nb[i].idx;
if (!used[q])
{
used[q] = true;
path.push_back(q);
dfs(q, t, dist + v[now].nb[i].dist, minDist, cost + v[now].nb[i].w, path, ans, v, used);
path.pop_back();
used[q] = false;
}
}
}
}
bool cmp(const pair<int, vector<int>>&a, const pair<int, vector<int>>&b)
{
return a.first < b.first;
}
int main(void)
{
int cityNum, edgeNum, s, t;
cin >> cityNum >> edgeNum >> s >> t;
vector<vertexNode> v(cityNum);
for (int i = 0; i < edgeNum; i++)
{
int a, b, d, w;
scanf("%d %d %d %d", &a, &b, &d, &w);
v[a].nb.push_back(neighbor(b, w, d));
v[b].nb.push_back(neighbor(a, w, d));
}
v[s].visited = true;
v[s].dist = 0;
while (1)
{
int p = -1;
for (int i = 0; i < v.size(); i++)
{
if (p == -1 && v[i].visited && !v[i].sured)
p = i;
else if (p != -1 && v[i].visited && !v[i].sured && v[p].dist > v[i].dist)
p = i;
}
if (p == -1) break;
v[p].sured = true;
if (v[t].sured) break;
for (int i = 0; i < v[p].nb.size(); i++)
{
int q = v[p].nb[i].idx;//获取邻居的编号
if (!v[q].sured && v[q].dist > v[p].dist + v[p].nb[i].dist)
{
v[q].visited = true;
v[q].dist = v[p].dist + v[p].nb[i].dist;
}
}
}
int minDist = v[t].dist;
vector<int> path(0);
path.push_back(s);
vector < pair<int, vector<int>>> ans(0);
vector<bool> used(cityNum, false);
dfs(s, t, 0, minDist, 0, path, ans, v, used);
sort(ans.begin(), ans.end(), cmp);
for (int i = 0; i < ans[0].second.size(); i++)
{
cout << ans[0].second[i] << " ";
}
cout << minDist << " " << ans[0].first;
return 0;
}
[/c]