PAT (Advanced Level) 1030 Travel Plan (Dijkstra)

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Dijkstra搜最短路，再dfs查费用最少的一条。模板题。 

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<climits>
using namespace std;
int INF = INT_MAX;
int n,m,s,d;

struct Edge{
	int next, dis, cost;
	Edge(){}
	Edge(int next, int dis, int cost) : next(next), dis(dis), cost(cost){}
};

vector<Edge> graph[505];
int dis[505];
int vis[505];
int min_distance;

void Dijkstra(){
	for(int i = 0; i < n; ++i){
		dis[i] = INF;
		vis[i] = 0;
	}
	dis[s] = 0;
	vis[s] = 1;
	for(int i = 0; i < graph[s].size(); ++i){
		dis[graph[s][i].next] = graph[s][i].dis;
	}
	int v, min_dis;
	while(1){
		min_dis = INF;
		v = -1;
		for(int i = 0; i < n; ++i){
			if(!vis[i] && min_dis > dis[i]){
				min_dis = dis[i];
				v = i;
			}
		}
		if(v == -1) break;
		vis[v] = 1;
		for(int i = 0; i < graph[v].size(); ++i){
			if(!vis[graph[v][i].next] && dis[v] + graph[v][i].dis < dis[graph[v][i].next]){
				dis[graph[v][i].next] = dis[v] + graph[v][i].dis;
			}
		}
	}
}
int min_cost = INF;
vector<int> final_path;
void dfs(int cur, int cost, int road, vector<int> path){
	if(road > min_distance) return;
	if(cur == d) {
		if(cost < min_cost){
			final_path = path;
			min_cost = cost;
		}
		return;
	}
	
	int mark;
	for(int i = 0; i < graph[cur].size(); ++i){
		if(!vis[graph[cur][i].next]){
			vis[graph[cur][i].next] = 1;
			path.push_back(graph[cur][i].next);
			dfs(graph[cur][i].next, cost + graph[cur][i].cost, road + graph[cur][i].dis,path);
			vis[graph[cur][i].next] = 0;
			path.pop_back();
		}
	}
}
int main(){
	scanf("%d%d%d%d",&n,&m,&s,&d);
	int a,b,dist,cost;
	for(int i = 0; i < m; ++i){
		scanf("%d%d%d%d",&a,&b,&dist,&cost);
		graph[a].push_back(Edge(b,dist,cost));
		graph[b].push_back(Edge(a,dist,cost));
	}
	Dijkstra();
	min_distance = dis[d];
	for(int i = 0; i < n; ++i) vis[i] = 0;
	vector<int> tmp_path;
	tmp_path.push_back(s);
	dfs(s,0,0,tmp_path);
	
	for(int i = 0; i < final_path.size(); ++i){
		cout << final_path[i] << " ";
	}
	cout << dis[d] << " " << min_cost << endl;
	return 0;
}