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## Discussion of Problem 1205. By the Underground or by Foot?

Why Dijkstra TLE#3 ??
Posted by Romko [Lviv NU] 13 Apr 2007 01:33
#include <iostream>
#include <cmath>
using namespace std;

struct point
{
double x;
double y;
static double dist(point &q, point &w)
{
return (sqrt((q.x - w.x)*(q.x - w.x)+(q.y - w.y)*(q.y - w.y)));
}
point():x(0),y(0){}
};
double A[210][210];

double mini(double a,double b)
{
if(a < b)
return a;
else
return b;
}
#define oo 10000;
bool S[210];
double B[210];
double C[210];
void Dijkstra(int a,int n)
{
for(int i = 1; i <= n; i++)
{
S[i] = 0;
C[i] = a;
B[i] = A[a][i];
}
B[a] = 0;S[a] = 1;C[a] = 0;
while(1)
{
int min = 10000;
int ind;
for(int i = 1; i <= n; i++)
{
if(!S[i])
{
if(B[i] < min)
{
min = B[i];
ind = i;
}
}
}
S[ind] = 1;
for(int k = 1; k <= n; k++)
{
if(!S[k])
{
if(B[k] > B[ind] + A[ind][k])
{
B[k]  = B[ind] + A[ind][k];
C[k] = ind;
}
}
}
bool tf = false;
for(int i = 1; i <= n; i++)
{
if(!S[i])
tf = true;
}
if(!tf)
return;
}
}
void output(int n)
{
int z = n;
int MT[210];
int i = 0;
do
{
MT[i] = z;
z = C[z];
i++;
}
while(z);
cout << i << ' ';
for(int j = i-1; j >= 0; j--)
cout << MT[j] << ' ';
}
int main()
{
int v1,v2;
cin >> v1 >> v2;
int n;
cin >> n;
point und[210];
for(int i = 1; i <= n; i++)
cin >> und[i].x >> und[i].y;
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= n; j++)
A[i][j] = oo;
}
int a = -1,b= -1;
while(a!=0 && b != 0)
{
cin >> a >> b;
A[b][a] = point::dist(und[b],und[a]);
A[a][b] = point::dist(und[a],und[b]);
}
point Q,W;
cin >> Q.x >> Q.y;
cin >> W.x >> W.y;
double min1 = 10000;
int item1,item2;
for(int i = 1; i <= n; i++)
{
if(point::dist(und[i],Q) < min1)
{
min1 = point::dist(und[i],Q);
item1 = i;
}
}
double min2 = 10000;
for(int i = 1; i <= n; i++)
{
if(point::dist(und[i],W) < min2)
{
min2 = point::dist(und[i],W);
item2 = i;
}
}
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= n; j++)
{
if((i != j) && (A[i][j] == 10000))
{
A[i][j] = point::dist(und[i],und[j])*v1*v2;
A[j][i] = point::dist(und[j],und[i])*v1*v2;
}
}
}
Dijkstra(item1,n);
double tmp = point::dist(Q,W);
if(tmp/v1 < B[item2]/v2 + min1/v1 + min2/v1)
{
printf("%.7lf\n0\n",tmp/v1);

}
else
{
double res = B[item2]/v2 + min1/v1 + min2/v1;
printf("%.7lf\n",res);
output(item2);
}

return 0;
}