POJ 2502 Subway （最短路）

Subway

题目链接：

http://acm.hust.edu.cn/vjudge/contest/122685#problem/L

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

Hint

##题意：

已知地铁的速度是40km/h, 步行速度是10km/h.
给出若干条地铁线路.
问从起点到终点的最快路径.


##题解：

很显然是建边然后求最短路.
需要注意的是：
1. 地铁线不一定是直线，所以同一地铁线上不相邻的点之间不能直接建"地铁边".
2. 还是由于不是直线，所以同一地铁线上不相邻的点之间步行可能要比地铁快.(因为步行是直线，地铁可能是曲线).


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 350
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n;

double value[maxn][maxn];

double dis[maxn];

bool vis[maxn];

int pre[maxn];

void dijkstra(int s) {

memset(vis, 0, sizeof(vis));

memset(pre, -1, sizeof(pre));

for(int i=0; i<=n; i++) dis[i] = inf;

dis[s] = 0;

for(int i=0; i<n; i++) {
    int p;
    double mindis = inf;
    for(int j=0; j<n; j++) {
        if(!vis[j] && dis[j]<mindis)
            mindis = dis[p=j];
    }
    vis[p] = 1;
    for(int j=0; j<n; j++) {
        if(dis[j] > dis[p]+value[p][j]) {
            dis[j] = dis[p] + value[p][j];
            pre[j] = p;
        }
    }
}

}

typedef pair<int,int> pii;

vector stop;

int first[maxn], last[maxn], lines;

int cnt;

int main(int argc, char const *argv[])

{

//IN;

for(int i=0; i<maxn; i++) for(int j=0; j<maxn; j++) value[i][j] = inf;
stop.clear(); lines = 1;

int a,b; cnt = 2; first[0]=0; last[0]=1;
cin >> a >> b;  stop.push_back(make_pair(a,b));
cin >> a >> b;  stop.push_back(make_pair(a,b));
value[0][1] = value[1][0] = 4.0*sqrt((double)((stop[0].first-stop[1].first)*(stop[0].first-stop[1].first) + (stop[0].second-stop[1].second)*(stop[0].second-stop[1].second)));
first[lines] = cnt;

while(scanf("%d %d",&a,&b) != EOF) {
    if(a==-1 && b==-1) {
        last[lines] = cnt-1;
        lines++;
        first[lines] = cnt;
        continue;
    }
    stop.push_back(make_pair(a,b));
    cnt++;
}

for(int l=1; l<lines; l++) {
    for(int i=first[l]; i<last[l]; i++) {
            int j = i + 1;
            double tmp = sqrt(double((stop[i].first-stop[j].first)*(stop[i].first-stop[j].first) + (stop[i].second-stop[j].second)*(stop[i].second-stop[j].second)));
            if(tmp < value[i][j]) value[i][j] = tmp;
            if(tmp < value[j][i]) value[j][i] = tmp;
    }
    for(int i=first[l]; i<=last[l]; i++) {
        for(int j=0; j<i; j++) {
            double tmp = sqrt(double((stop[i].first-stop[j].first)*(stop[i].first-stop[j].first) + (stop[i].second-stop[j].second)*(stop[i].second-stop[j].second)));
            tmp *= 4.0;
            if(tmp < value[i][j]) value[i][j] = tmp;
            if(tmp < value[j][i]) value[j][i] = tmp;
        }
    }
}

n = cnt;
dijkstra(0);

double ans = (dis[1]*3.0/2000.0);

printf("%.0f\n", ans);

return 0;

}