PAT.1030 Travel Plan - Dijikstra

最新推荐文章于 2025-06-04 23:52:10 发布

王马软

最新推荐文章于 2025-06-04 23:52:10 发布

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分类专栏：做一手PAT 图文章标签：深度优先 c++ 算法 PAT Dijikstra

本文链接：https://blog.youkuaiyun.com/weixin_43662405/article/details/123423749

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本文介绍了如何使用Dijkstra算法求解旅行计划中所有可能的最短路线，并通过深度优先搜索找到成本最低的路径。通过实例演示了如何在城市间计算最短距离和最少花费，适合理解图算法在实际问题中的应用。

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PAT.1030 Travel Plan - Dijikstra

题目链接
思路和PAT.1018基本是一样的，先Dijikstra求所有可能的最短路，然后dfs求所有最短路里代价最小的一条。

题解

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;

struct city{
    int num;
    int dis;
    bool operator < (const city &a) const{
        if(dis > a.dis) return true;
        else return false;
    }
};

int cityCnt,roadCnt,start,s,d,l,c,dest,dis[505][505],cost[505][505],vis[505],minDis[505],minCost = 0x3f3f3f3f;
vector<int> neighbors[505];
vector<int> preCity[505];
vector<int> res;
priority_queue<city> citys;

void dfs(int city,int currentCost,vector<int> path){
    if(city == start){
        if(currentCost < minCost){
            minCost = currentCost;
            res = path;
        }
        return;
    }
    for(int pc : preCity[city]){
        path.push_back(pc);
        dfs(pc,currentCost + cost[pc][city],path);
        path.pop_back();
    }
}

int main(){
    cin>>cityCnt>>roadCnt>>start>>dest;
    memset(minDis,0x3f3f3f3f,sizeof(minDis));
    for(int i = 0 ; i < roadCnt ; ++i){
        cin>>s>>d>>l>>c;
        dis[s][d] = dis[d][s] = l;
        cost[s][d] = cost[d][s] = c;
        neighbors[s].push_back(d);
        neighbors[d].push_back(s);
    }
    citys.push((city){start,0});
    while(!citys.empty()){
        city currentCity = citys.top();
        citys.pop();
        if(vis[currentCity.num] == 1) continue;
        vis[currentCity.num] = 1;
        for(int nextCity : neighbors[currentCity.num]){
            if(dis[currentCity.num][nextCity] + currentCity.dis < minDis[nextCity]){
                minDis[nextCity] = dis[currentCity.num][nextCity] + currentCity.dis;
                preCity[nextCity].clear();
                preCity[nextCity].push_back(currentCity.num);
                citys.push((city){nextCity,minDis[nextCity]});
            }else if(dis[currentCity.num][nextCity] + currentCity.dis == minDis[nextCity]){
                preCity[nextCity].push_back(currentCity.num);
            }
        }
    }
    //find the path with least cost;
    dfs(dest,0,{dest});
    for(auto itr = res.rbegin() ; itr != res.rend() ; ++itr){
        cout<<*itr<<' ';
    }
    cout<<minDis[dest]<<' '<<minCost;
}