poj2502題解(建圖+最短路)
阿新 • • 發佈:2018-10-01
double 思路 ios getc clu from scanf ont 解題思路
題意
給起點和終點的坐標,然後給出多條地鐵每一站的坐標,每站地鐵只能到相鄰的地鐵站,地鐵的速度是40km/h,人行走的速度是10km/h,求起點到終點的最小時間(給出的坐標單位是m,最後求的時間單位是分鐘)
解題思路
這題關鍵點在與建圖,把圖建好後跑dijkstra就很簡單了,之前一直wa是想復雜了,以為要考慮起點和終點在地鐵站的情況和不同線路的地鐵站相交的情況,後來想不需要這麽麻煩,同一條線路相鄰的40,不同的10就可以了,相交的話,以10km/h走時間花費是0。
AC代碼
#include<vector> #include<algorithm> #include<cstdio> #include<iostream> #include<set> #include<cmath> #include<cstring> #include<map> #include<queue> using namespace std; typedef long long ll; typedef pair<double,int> pii; typedef pair<int,pii> PII; const int inf = 1<<30; const int maxn = 1e5+5; struct Node { int x,y; Node(){} Node(int a,int b):x(a),y(b){} }; Node beg,en; vector<Node> node; Node D[maxn]; struct Edge { int from,to; double val; double cost; }; vector<Edge> G[maxn]; double getval(Node &a,Node& b) { return sqrt((double)(abs(a.x-b.x)*abs(a.x-b.x)+abs(a.y-b.y)*abs(a.y-b.y))); } double getcost(double cnt,bool issubway) { double ans = 0; if(issubway){ ans = cnt/40000 * 60; }else{ ans = cnt/10000 * 60; } return ans; } double dis[maxn]; bool vis[maxn]; int cnt = 0; double dijkstra() { memset(vis,0,sizeof(vis)); fill(dis,dis+maxn,inf); priority_queue<pii,vector<pii>,greater<pii> > Q; Q.push(pii(0,cnt)); while(!Q.empty()){ pii t = Q.top(); Q.pop(); int u = t.second; double d = t.first; if(vis[u])continue; vis[u] = 1; for(int i=0;i<G[u].size();i++){ Edge e = G[u][i]; if(e.cost + d < dis[e.to]){ dis[e.to] = e.cost + d; Q.push(pii(dis[e.to],e.to)); } } } return dis[cnt+1]; } int main(int argc, char const *argv[]) { scanf("%d%d",&beg.x,&beg.y); scanf("%d%d",&en.x,&en.y); int x,y; int k = 0; Edge e; while(~scanf("%d%d",&x,&y)){ if(x==-1&&y==-1){ for(int i=k;i<cnt-1;i++){ e.val = getval(D[i], D[i+1]); e.cost = getcost(e.val, 1); e.from = i; e.to = i+1; G[e.from].push_back(e); swap(e.from, e.to); G[e.from].push_back(e); } k = cnt; continue; } D[cnt] = Node(x,y); cnt++; } D[cnt] = beg; D[cnt+1] = en; for(int i=0;i<cnt+2;i++){ for(int j=i+1;j<cnt+2;j++){ e.val = getval(D[i], D[j]); e.cost = getcost(e.val, 0); e.from = i; e.to = j; G[e.from].push_back(e); swap(e.from, e.to); G[e.from].push_back(e); } } double ans = dijkstra(); printf("%d\n",(int)(ans+0.5)); return 0; }
poj2502題解(建圖+最短路)