[PAT]1030 Travel Plan——Dijkstra+DFS

1030 Travel Plan （30 分）

一、原题

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

二、代码

以下基于dijkstra：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int maxv=510;
const int inf=100000000;

int g[maxv][maxv],d[maxv],c[maxv],cost[maxv][maxv],pre[maxv];
bool vis[maxv]={false};
int n,m,s,de;

void dijkstra(int s)
{
    fill(d,d+maxv,inf);
    fill(c,c+maxv,inf);
    for(int i=0;i<n;i++) pre[i]=i; //学习0：赋并查集初值

    d[s]=0;
    c[s]=0;
    for(int i=0;i<n;i++)
    {
        int u=-1,mm=inf;
        for(int j=0;j<n;j++)
        {
            if(vis[j]==false && d[u]<mm)
            {
                u=j;
                mm=d[j];
            }
        }
        if(u==-1) return;
        vis[u]=true;
        for(int j=0;j<n;j++)
        {
            if(vis[j]==false && g[u][j]<inf)
            {
                if(d[u]+g[u][j]<d[j])
                {
                    d[j]=d[u]+g[u][j];
                    c[j]=c[u]+cost[u][j];
                    pre[j]=u;
                }
                else if(d[u]+g[u][j]==d[j]){
                        //学习一：是当有更优时，因为题中说了路径唯一！
                    if(c[u]+cost[u][j]<c[j]) {c[j]=c[u]+cost[u][j];pre[j]=u;}
                }
            }
        }
    }

}
void dfs(int v)//学习二：打印路径
{
    if(v==s)
    {
        printf("%d ",v);
        return;
    }
    dfs(pre[v]);
    printf("%d ",v);
}


int main()
{
   scanf("%d %d %d %d",&n,&m,&s,&de);
    fill(g[0],g[0]+maxv*maxv,inf);//学习三：初始化图
    for(int i=0;i<m;i++)
    {
        int c1,c2;
        scanf("%d%d",&c1,&c2);
        scanf("%d",&g[c1][c2]);
        g[c2][c1]=g[c1][c2];
        scanf("%d",&cost[c1][c2]);
        cost[c2][c1]=cost[c1][c2];
    }
    dijkstra(s);
    dfs(de);
    printf("%d %d",d[de],c[de]);
    return 0;
}

AC版：(上面代码只能部分AC，没找出问题，烦劳哪位大佬看出来了告知下）

#include<bits/stdc++.h>
#include<cmath>
 
#define mem(a,b) memset(a,b,sizeof a)
#define ssclr(ss) ss.clear(), ss.str("")
#define INF 0x3f3f3f3f
#define MOD 1000000007
 
using namespace std;
 
typedef long long ll;
 
const int maxn=510;
 
int n,m,s,d;
int mp[maxn][maxn], cost[maxn][maxn];
int vis[maxn], dis[maxn], cst[maxn], pre[maxn];
vector<int> vec;
 
void init()
{
    mem(vis,0), mem(cst,0), mem(dis,INF), mem(pre,-1);
    mem(mp,INF), mem(cost,INF);
}
 
int dijkstra(int s)
{
    dis[s]=0;
    while(1)
    {
        int mi=INF,s=-1;
        for(int j=0;j<n;j++)
        {
            if(!vis[j] && dis[j]<mi)
                mi=dis[j], s=j;
        }
 
        if(s==d) return 1;
        if ( s==-1 ) return 0;
        vis[s] = 1;
 
        for(int j=0;j<n;j++)
        {
            if(!vis[j] && mi+mp[s][j]<dis[j])
            {
                pre[j]=s;
                dis[j]=mi+mp[s][j];
                cst[j]=cst[s]+cost[s][j];
            }
            else if(!vis[j] && mi+mp[s][j]==dis[j] && cst[j]>cst[s]+cost[s][j])
                cst[j]=cst[s]+cost[s][j], pre[j]=s;
        }
    }
}
 
int main()
{
    init();
    int a,b,u,v;
    scanf("%d%d%d%d",&n,&m,&s,&d);
    for(int i=0;i<m;i++)
    {
        scanf("%d%d%d%d",&u,&v,&a,&b);
        mp[u][v]=mp[v][u]=a;
        cost[u][v]=cost[v][u]=b;
    }
    if(dijkstra(s))
    {
        vec.clear();
        int h=d;
        while(h!=-1)
        {
            vec.push_back(h);
            h=pre[h];
        }
 
        for(int i=vec.size()-1;i>=0;i--) printf("%d ",vec[i]);
        printf("%d %d\n",dis[d],cst[d]);
    }
    else puts("-1");
 
    return 0;
}

三、小结

参考
DFS+Dijkstra

void dfs(int v)
{
    temppath.push_back(v);
   if(v == st) {
        int tempcost=0; //记录路径花费
        for(int i = temppath.size() - 1; i >= 0; i--) {
            int id=temppath[i],idnext=temppath[i-1];
            tempcost+=cost[id][idnext];
        }
        if(tempcost<mincost)
        {
            mincost=tempcost;
            path=temppath;
        }
        temppath.pop_back();
        return ;
    }
        for(int i=0;i<pre[v].size();i++)
        {
            dfs(pre[v][i]);
        }
        temppath.pop_back();
}