hdu1245 两个权值的最短路

题意:
      求s到t的最短路,如果路径相同求那么要求另一个权值尽可能的小.
思路:

      水题,就是spfa的比较那个地方多了一个可以更新的机会,当(s_x[xin] > s_x[tou] + E[k].cost || s_x[xin] == s_x[tou] + E[k].cost && s_t[xin] > s_t[tou] + 1) 时更新就行了..

#include<stdio.h>
#include<string.h>
#include<queue>
#include<math.h>

#define N_node 100 + 5
#define N_edge 20000 + 500
#define INF 1000000000

using namespace std;

typedef struct
{
   int to ,next;
   double cost;
}STAR;

typedef struct
{
   double x ,y;
}NODE;

STAR E[N_edge];
NODE node[N_node];
int list[N_node] ,tot;
double s_x[N_node];
int s_t[N_node];

void add(int a ,int b ,double c)
{
   E[++tot].to = b;
   E[tot].cost = c;
   E[tot].next = list[a];
   list[a] = tot;
}

double abss(double x)
{
   return x > 0 ? x : -x;
}
double minn(double x ,double y)
{
   return x < y ? x : y;
}

void SPFA(int s ,int n)
{
   for(int i = 0 ;i <= n ;i ++)
   s_x[i] = INF ,s_t[i] = INF;
   int mark[N_node] = {0};
   mark[s] = 1;
   s_x[s] = s_t[s] = 0;
   queue<int>q;
   q.push(s);
   while(!q.empty())
   {
      int xin ,tou;
      tou = q.front();
      q.pop();
      mark[tou] = 0;
      for(int k = list[tou] ;k ;k = E[k].next)
      {
         xin = E[k].to;
         if(s_x[xin] > s_x[tou] + E[k].cost || abss(s_x[xin] - s_x[tou] + E[k].cost) < 1e-6 && s_t[xin] > s_t[tou] + 1)
         {
            s_x[xin] = s_x[tou] + E[k].cost;
            s_t[xin] = s_t[tou] + 1;
            if(!mark[xin])
            {
               mark[xin] = 1;
               q.push(xin);
            }
         }
      }
   }
   return ;
}

int main ()
{
   int n ,i ,j;
   double d;
   while(~scanf("%d %lf" ,&n ,&d))
   {
      for(i = 1 ;i <= n ;i ++)
      scanf("%lf %lf" ,&node[i].x ,&node[i].y);
      memset(list ,0 ,sizeof(list));
      tot = 1;
      for(i = 1 ;i <= n ;i ++)
      for(j = i + 1 ;j <= n ;j ++)
      {
         double dis = pow(node[i].x - node[j].x ,2.0) + pow(node[i].y - node[j].y,2.0);
         if(dis <= d * d) 
         {
            add(i ,j ,sqrt(dis));
            add(j ,i ,sqrt(dis));
         }
      }
      
      int s = 0 ,t = n + 1;
      for(i = 1 ;i <= n ;i ++)
      {
         double dis = pow(node[i].x,2.0) + pow(node[i].y ,2.0);
         if(pow(d + 7.5 ,2.0) >= dis)
         add(s ,i ,sqrt(dis) - 7.5);
         dis = minn(50 - abss(node[i].x) ,50 - abss(node[i].y));
         if(dis <= d) add(i ,t ,dis);
      }
      if(d >= 50 - 7.5)
      add(s ,t ,50 - 7.5);
      SPFA(s ,t);
      if(s_x[t] == INF)
      printf("can't be saved
");
      else
      printf("%.2lf %d
" ,s_x[t] ,s_t[t]);
   }
   return 0;
}