1030 Travel Plan （30 分）--PAT甲级

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本文深入探讨了如何使用Dijkstra算法结合深度优先搜索(DFS)来寻找两点间花费最少的最短路径。通过实例讲解了算法的具体实现过程，包括初始化距离矩阵、成本矩阵，运用Dijkstra算法更新最短路径，并通过DFS遍历所有可能路径找到最小成本路径。

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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:

Sample Output:

0 2 3 3 40

求花费最少的最短路径。
直接dijkstra+dfs吧，同样是30分，1111 Online Map （30 分）就比这道麻烦多了。

#include <iostream>
#include<cstdio>
#include<cmath>
#include <algorithm>
#include<vector>
using namespace std;
const int maxn = 505;
const int inf = 0x3fffffff;
int n,first,last;
int dist[maxn][maxn];
int cost[maxn][maxn];
int d[maxn];
bool visit[maxn];
vector<vector<int> >pre;
vector<int>path;
int mincost=inf;

void dijkstra(int s){
    fill(d,d+n,inf);
    d[s]=0;
    for(int i=0;i<n;i++){
        int u=-1,MIN=inf;
        for(int j=0;j<n;j++){
            if(visit[j]==false && d[j]<MIN){
                MIN=d[j];
                u=j;
            }
        }
        if(u==-1) return;
        visit[u]=true;
        for(int v=0;v<n;v++){
            if(visit[v]==false && dist[u][v]!=0){
                if(dist[u][v]+d[u]<d[v]){
                    d[v]=dist[u][v]+d[u];
                    pre[v].clear();
                    pre[v].push_back(u);
                }
                else if(dist[u][v]+d[u]==d[v]){
                    pre[v].push_back(u);
                }
            }
        }
    }
}

void dfs(int v,vector<int>temp,int cst){
    temp.push_back(v);
    if(v==first){
        if(cst<mincost){
            mincost=cst;
            path=temp;
        }
        return;
    }
    else{
        for(int i=0;i<pre[v].size();i++){
            dfs(pre[v][i],temp,cst+cost[pre[v][i]][v]);
        }
    }
}

int main()
{
    int m;
    scanf("%d %d %d %d",&n,&m,&first,&last);
    pre.resize(n);
    for(int i=0;i<m;i++){
        int a,b,c,d;
        scanf("%d %d %d %d",&a,&b,&c,&d);
        dist[a][b]=c;
        dist[b][a]=c;
        cost[a][b]=d;
        cost[b][a]=d;
    }
    dijkstra(first);
    vector<int>t;
    dfs(last,t,0);
    for(int i=path.size()-1;i>=0;i--){
        printf("%d ",path[i]);
    }
    printf("%d %d\n",d[last],mincost);
    return 0;
}