最短路练习12/poj/2502 Subway dijkstar，spfa，floyd都可以解决

题目链接：http://poj.org/problem?id=2502

Subway

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9766 Accepted: 3163

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 10000 200 5000 200 7000 200 -1 -1 2000 600 5000 600 10000 600 -1 -1

Sample Output

21

这道题做的时间不短，花了将近4小时，自己的代码错误一直没找出来。看了题解也找了2小时才找出错误。

发现自己思考问题不全面，并且一旦入坑就很难跳出。（无法摆脱自己的定向思维）（这是个大问题）

题意：现在学生要去学校，学生和学校的坐标第一行给出，接下来每一行都代表一条地铁的轨道，直到-1，-1表明这条地铁结束，地铁站的每个坐标已给出。求小明到达学校的最短时间。

思路难点：建图非常困难，靠考虑很多细节，比如：地铁的第一站，要与前面出现的所有点都要求出距离（求出步行时间）。地铁的其他站（二站，三站，四站等等），与他的前一站求出（火车行走时间），并且与（除了他的前一站外的）所有坐标求出步行时间，地铁（不相邻的站）的路线并不一定是直的。

因为有可能出现这种情况站1到站3的步行时间可能要比乘地铁时间更短。所以只有相邻的两站我们才用地铁速度，其他都用步行速度。

还有一个挺坑的地方，样例都不能让咱们轻松输出。你必须把多组输入改一下才可以，改成k=8，while（k--）{  }；这样就能试样例了，提交前不要忘了该成多组输入哦。

dijkstar，spfa 是我自己写的代码，比较低调。最后的floyd是搜的博客，比较高大上。

试了各种方法做这道题。dijkstar  AC代码：

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<vector>#include<map>#include<string>#define LL long long#define eps 1e-8#define N 3005#define M 100005using namespace std;const int mod = 1e7+7;const double inf = 1000000000.0;const int maxx = 1e6 +10;struct node{    int x,y;} p[N];int cnt,sum,f;double dis[N],fx;int vis[N];double ma[N][N];void dijkstra(){    for(int i=0;i<cnt;i++)        dis[i]=ma[1][i];     memset(vis,0,sizeof(vis));     vis[1]=1;     dis[1]=0;     double sum;     int point;     for(int i=1;i<cnt;i++)     {         sum=inf;         point =-1;         for(int j=1;j<cnt;j++)         {             if(!vis[j]&&dis[j]<sum)             {                 sum=dis[j];                 point =j;             }         }         if(point!=-1)         {             vis[point]=1;             for(int j=1;j<cnt;j++)             {                 if(dis[point]+ma[point][j]<dis[j])                 {                     dis[j]=dis[point ]+ma[point ][j];                 }             }         }     }}int main(){    cnt=3;    sum=0;    for(int i=0; i<240; i++)        for(int j=0; j<240; j++)            ma[i][j]=i==j?0.0:inf;    scanf("%d%d%d%d",&p[1].x,&p[1].y,&p[2].x,&p[2].y);    ma[2][1]=ma[1][2]=sqrt((double)(p[1].x-p[2].x)*(double)(p[1].x-p[2].x)+(double)(p[1].y-p[2].y)*(double)(p[1].y-p[2].y))*60.0/10000.0;    while(~scanf("%d%d",&p[cnt].x,&p[cnt].y))    {        if(p[cnt].x==-1&&p[cnt].y==-1)        {            sum=0;            continue;        }        for(int i=1; i<cnt-sum; i++)//注意这里的sum的值用的非常巧妙，注意体会sum==0或1的用意            ma[i][cnt]=ma[cnt][i]=sqrt((double)(p[cnt].x-p[i].x)*(double)(p[cnt].x-p[i].x)+(double)(p[cnt].y-p[i].y)*(double)(p[cnt].y-p[i].y))*3.0/500.0;        if(sum)            ma[cnt][cnt-1]=ma[cnt-1][cnt]=sqrt((double)(p[cnt].x-p[cnt-1].x)*(double)(p[cnt].x-p[cnt-1].x)+(double)(p[cnt].y-p[cnt-1].y)*(double)(p[cnt].y-p[cnt-1].y))*3.0/2000.0;        cnt++;sum=1;    }    dijkstra();    printf("%.0lf\n",dis[2]);}

spfa  AC代码：

#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<vector>#include<map>#include<string>#define LL long long#define eps 1e-8#define N 30005#define M 100005using namespace std;const int mod = 1e7+7;const double inf = 1000000000.0;const int maxx = 1e6 +10;struct node {int x,y;}p[N];int cnt,sum,f;double dis[N];int vis[N];double ma[250][205];void spfa(){    memset(vis,0,sizeof(vis));    for(int i=0;i<240;i++)        dis[i]=inf;    queue<int>q;    dis[1]=0;    vis[1]=1;    q.push(1);    while(!q.empty())    {        f=q.front();        q.pop();        vis[f]=0;        for(int i=1;i<cnt;i++)        {            if(dis[i]>dis[f]+ma[f][i])            {                dis[i]=dis[f]+ma[f][i];                if(!vis[i])                {                    vis[i]=1;                    q.push(i);                }            }        }    }}int main(){    cnt=3;sum=0;    for(int i=0;i<240;i++)      for(int j=0;j<240;j++)        ma[i][j]=i==j?0.0:inf;    scanf("%d%d%d%d",&p[1].x,&p[1].y,&p[2].x,&p[2].y);    ma[2][1]=ma[1][2]=sqrt((double)(p[1].x-p[2].x)*(double)(p[1].x-p[2].x)+(double)(p[1].y-p[2].y)*(double)(p[1].y-p[2].y))*60.0/10000.0;    while(~scanf("%d%d",&p[cnt].x,&p[cnt].y))    {        if(p[cnt].x==-1&&p[cnt].y==-1)        {            sum=0;            continue;        }        for(int i=1;i<cnt-sum;i++)            ma[i][cnt]=ma[cnt][i]=sqrt((double)(p[cnt].x-p[i].x)*(double)(p[cnt].x-p[i].x)+(double)(p[cnt].y-p[i].y)*(double)(p[cnt].y-p[i].y))*60.0/10000.0;           if(sum)             ma[cnt-1][cnt]=ma[cnt][cnt-1]=sqrt((double)(p[cnt].x-p[cnt-1].x)*(double)(p[cnt].x-p[cnt-1].x)+(double)(p[cnt].y-p[cnt-1].y)*(double)(p[cnt].y-p[cnt-1].y))*60.0/40000.0;        cnt++;sum=1;    }    spfa();    printf("%d\n",(int)(dis[2]+0.5));}

flody  AC代码：

#include <iostream>  #include <cstring>  #include <deque>  #include <cmath>  #include <queue>  #include <stack>  #include <list>  #include <map>  #include <set>  #include <string>  #include <vector>  #include <cstdio>  #include <bitset>  #include <algorithm>  using namespace std;  #define Debug(x) cout << #x << " " << x <<endl  #define Memset(x, a) memset(x, a, sizeof(x))  const int INF = 0x3f3f3f3f;  typedef long long LL;  typedef pair<int, int> P;  #define FOR(i, a, b) for(int i = a;i < b; i++)  #define MAX_N 2520  int n;  double G[MAX_N][MAX_N];  struct Point{      int x;      int y;  }p[MAX_N];  double walk(int i, int j){      double l = sqrt((double)(p[i].x - p[j].x) * (p[i].x - p[j].x) + (double)(p[i].y - p[j].y) * (p[i].y - p[j].y));      return  l * 60 / 10000.0;  }  double subway(int i, int j){      double l = sqrt((double)(p[i].x - p[j].x) * (p[i].x - p[j].x) + (double)(p[i].y - p[j].y) * (p[i].y - p[j].y));      return l * 60.0 / 40000.0;  }  void solve(){      for(int k = 1; k <= n; k++){          for(int i = 1;i <= n; i++){              for(int j = 1;j <= n; j++){                  G[i][j] = min(G[i][j], G[i][k] + G[k][j]);              }          }      }  }     int main(){      //freopen("in.cpp", "r", stdin);      scanf("%d%d%d%d", &p[1].x, &p[1].y, &p[2].x, &p[2].y);      G[1][2] = G[2][1] = walk(1, 2);      int j = 0;n = 3;      int a, b;      while(scanf("%d%d",&a,&b) != EOF){            if(a == -1 && b == -1)                 j = 0;            else{                 p[n].x = a, p[n].y = b;                for(int i = 1;i < n - j; i++){                     G[i][n] = G[n][i] = walk(i,n);                }                 for(int i = n-j; i < n; i++){                    G[i][n] = G[n][i] = subway(i,n);                }                j = 1, n++;             }         }          n--;      solve();      printf("%.0f\n", G[1][2]);      return 0;  }