Subway
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14634 | Accepted: 4718 |
题目链接：http://poj.org/problem?id=2502
Description:
You have just moved from a quiet Waterloo neighbourhood to a big, noisy city. Instead of getting to ride your bike to school every day, you now get to walk and take the subway. Because you don't want to be late for class, you want to know how long it will take you to get to school.
You walk at a speed of 10 km/h. The subway travels at 40 km/h. Assume that you are lucky, and whenever you arrive at a subway station, a train is there that you can board immediately. You may get on and off the subway any number of times, and you may switch between different subway lines if you wish. All subway lines go in both directions.
Input:
Input consists of the x,y coordinates of your home and your school, followed by specifications of several subway lines. Each subway line consists of the non-negative integer x,y coordinates of each stop on the line, in order. You may assume the subway runs in a straight line between adjacent stops, and the coordinates represent an integral number of metres. Each line has at least two stops. The end of each subway line is followed by the dummy coordinate pair -1,-1. In total there are at most 200 subway stops in the city.
Output:
Output is the number of minutes it will take you to get to school, rounded to the nearest minute, taking the fastest route.
Sample Input:
0 0 10000 1000 0 200 5000 200 7000 200 -1 -1 2000 600 5000 600 10000 600 -1 -1
Sample Output:
21
题意：
在二维平面上，给出起点和终点的坐标。另外还有多条地铁线路，给出每条线路有哪些站。
如果坐地铁的速度是40km/h，而走路的速度为10km/h，现在问从起点到终点最少花费多少时间。
题解：
直接模拟建边就行了，注意要换算一下单位，因为最后问的是分钟。
有个需要注意的地方就是地铁站之间也可以走路。另外结果四舍五入！！
代码如下：
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <queue> #include <vector> #include <cmath> #include <map> #define INF 0x3f3f3f3f using namespace std; typedef long long ll; const int N = 805; int x,y,X,Y,cnt,num,s,t; map <ll,map<ll,int> > p; double d[N]; int vis[N],head[N]; struct node{ int x,y; }; vector <node> vec[N]; double dis(int X1,int X2,int Y1,int Y2){ return sqrt((double)(X1-X2)*(X1-X2)+(double)(Y1-Y2)*(Y1-Y2)); } double T(double d,int op){ if(op==1) return 6*d/1000.0; else return 6*d/4000.0; } struct Edge{ int u,v,next ; double w; }e[N*N<<2]; int tot; struct Node{ double d; int u; bool operator < (const Node &A)const{ return d>A.d; } }; void adde(int u,int v,double w){ e[tot].v=v;e[tot].w=w;e[tot].next=head[u];head[u]=tot++; } void Dijkstra(int s){ priority_queue <Node> q; for(int i=0;i<=t;i++) d[i]=INF; memset(vis,0,sizeof(vis));d[s]=0; Node now; now.d=0;now.u=s; q.push(now); while(!q.empty()){ Node cur = q.top();q.pop(); int u=cur.u; if(vis[u]) continue ; vis[u]=1; for(int i=head[u];i!=-1;i=e[i].next){ int v=e[i].v; if(d[v]>d[u]+e[i].w){ d[v]=d[u]+e[i].w; now.d=d[v];now.u=v; q.push(now); } } } } int main(){ cin>>x>>y>>X>>Y; int xx,yy; memset(head,-1,sizeof(head)); while(scanf("%d",&xx)!=EOF){ scanf("%d",&yy); cnt++; node cur; cur.x=xx;cur.y=yy; vec[cnt].push_back(cur); while(1){ scanf("%d%d",&xx,&yy); if(xx==-1 && yy==-1) break ; cur.x=xx;cur.y=yy; vec[cnt].push_back(cur); } } s=0;t=300; for(int i=1;i<=cnt;i++){ for(int j=0;j<vec[i].size();j++){ node cur = vec[i][j]; if(p[cur.x][cur.y]==0) p[cur.x][cur.y]=++num; adde(s,p[cur.x][cur.y],T(dis(0,cur.x,0,cur.y),1)); adde(p[cur.x][cur.y],t,T(dis(cur.x,X,cur.y,Y),1)); } } adde(s,t,T(dis(x,X,y,Y),1)); for(int i1=1;i1<=cnt;i1++){ for(int j1=0;j1<vec[i1].size();j1++){ int len1 = vec[i1].size(); node cur1 = vec[i1][j1]; if(j1>0){ node cur2 = vec[i1][j1-1]; adde(p[cur1.x][cur1.y],p[cur2.x][cur2.y],T(dis(cur1.x,cur2.x,cur1.y,cur2.y),2)); } if(j1<len1-1){ node cur2 = vec[i1][j1+1]; adde(p[cur1.x][cur1.y],p[cur2.x][cur2.y],T(dis(cur1.x,cur2.x,cur1.y,cur2.y),2)); } for(int i2=1;i2<=cnt;i2++){ for(int j2=0;j2<vec[i2].size();j2++){ node cur2 = vec[i2][j2]; adde(p[cur1.x][cur1.y],p[cur2.x][cur2.y],T(dis(cur1.x,cur2.x,cur1.y,cur2.y),1)); } } } } Dijkstra(s); cout<<(int)(d[t]+0.5); return 0; }