Minimum cost c++ detailed explanation

minimum cost

topic description

at $Among\; n$ individuals, the bank accounts of certain individuals can transfer money to each other. Fees for transferring money between these people vary. Given that a few percent of the handling fee needs to be deducted from the transfer amount when transferring money between these people, ask$How\; much\; money\; does\; A$ need at least to make BBafter the transfer$B$ receives$100$ yuan.

input format

The first line enters two positive integers $n,m$ , represent the logarithm of the total number of people and the number of people who can transfer money to each other, respectively.

Below mmEnter three positive integers x , y , zx,y,zin each line of$m\; lines$$x,y,z$ , means the label isxxperson$x$ and label yyTransfer between$y$$z\%$ commission$(z<100)$。

The last line enters two positive integers $A,B.$ _ Data$A$ and$B$ can transfer funds directly or indirectly.

output format

Export $A$ makes$B$ received100 100A minimum total fee of $$100\; is\; required.$Accurate to 8after the decimal point$8$ bits.

sample

sample input

3 3                                     
1 2 1
2 3 2
1 3 3
1 3

sample output

103.07153164

hint

$1\le n\le 2000,m\le 100000$。

solution

As can be seen from the meaning of the title, the label is xxperson$x$ and label yyTransfer between$y$$z\%$ commission$(z<100)$ .
That is to say:
the label isxxperson$x$$When\; people\; of\; y$ transfer money to each other, only the original money will be received$(100-z)\%$
and after the intermediary transfer, the money received is the original$(100-z_1)\%\times (100-z_2)\%$
then we can run a Dijkstra fix.

the code

#include <bits/stdc++.h>
using namespace std;
int n,m,x,y,z,s,t,p[2200];
double f[2200][2200],c[2200];
int main()
{
    
    
	cin >>n >>m;
	for (int i=1;i<=m;i++)
	{
    
    
	    cin >>x >>y >>z;
	    f[x][y]=f[y][x]=(100.0-z)/100.0;
	}
	cin >>s >>t;
	p[s]=1;
	for (int i=1;i<=n;i++)
	    c[i]=f[s][i];
	for (int i=1;i<n;i++)
	{
    
    
		int k=0;
		double ma=0.0;
		for (int j=1;j<=n;j++)
		    if (p[j]==0 && ma<c[j])
		    {
    
    
		    	k=j;
		    	ma=c[j];
		    }
		p[k]=1;
		for (int j=1;j<=n;j++)
		    if (p[j]==0 && c[k]*f[k][j]>c[j])
		       c[j]=c[k]*f[k][j];
	}
	printf("%.8lf",100.0/c[t]);
	return 0;
}