fast 就是 cst ems names mis center lines getting

Subway

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14634		Accepted: 4718

題目鏈接：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

題意：

在二維平面上，給出起點和終點的坐標。另外還有多條地鐵線路，給出每條線路有哪些站。

如果坐地鐵的速度是40km/h，而走路的速度為10km/h，現在問從起點到終點最少花費多少時間。

題解：

直接模擬建邊就行了，註意要換算一下單位，因為最後問的是分鐘。

有個需要註意的地方就是地鐵站之間也可以走路。另外結果四舍五入！！

代碼如下：

#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;
}

POJ2502:Subway（最短路）