35分观察到有六七个点ai0我们选择直接无视其他点假装小z进入景区时所有道路都能通行了那么问题就转换成一个简单的广搜了用一个队列一层一层的把景点压入当到了终点时就是最省时间的了。#includebits/stdc.husingnamespacestd;intn,m,k;vectorpairint,intg[10005];boolvis[10005][10005];voidbfs(){queuepairint,intq;q.push({1,0});while(!q.empty()){intuq.front().first;inttq.front().second;q.pop();if(unt%k0){coutt;return;}if(vis[u][t%k])continue;vis[u][t%k]1;for(inti0;ig[u].size();i){q.push({g[u][i].first,t1});}}cout-1;return;}intmain(){cinnmk;while(m--){intu,v,w;cinuvw;g[u].push_back({v,w});}bfs();return0;}正解单源最短路可以用dijkstra来做。(这里给大家一个学迪杰斯特拉的网址我老师推荐的https://www.bilibili.com/video/BV1uT4y1p7Jy/?spm_id_from333.337.search-card.all.clickvd_sourceb59c08196148ea3ea8a331da551bf5a8总而言之dijkstra可以在中间添加路开通的时间判断能否走下一步找最短时间。#includebits/stdc.husingnamespacestd;intn,m,k,ans1e9;vectorvectorpairint,intg;boolvis[10005][105];voiddijkstra(){priority_queuepairint,int,vectorpairint,int,greaterpairint,intq;//建立小根堆维护最短路径q.push({0,1});while(!q.empty()){pairint,intcurq.top();q.pop();inttcur.first;intucur.second;if(unt%k0){ansmin(ans,t);}if(vis[u][t%k])continue;vis[u][t%k]1;for(inti0;ig[u].size();i){intvg[u][i].first;intwg[u][i].second;intntt1;if(ntw){nt(w-ntk)/k*k;}q.push({nt,v});}}}intmain(){cinnmk;g.resize(n1);for(inti1;im;i){intu,v,w;cinuvw;g[u].push_back({v,w});}dijkstra();if(ans1e9)cout-1;elsecoutans;return0;}