Minimum cost (multiplication shortest path)

Title

Given an undirected graph, the edge weight is $w_i(0\le w_i\le 1)$ . Define the distance between two points as the product of all the values ​​on the path. Find the longest way between two points.

Ideas

Suppose the distance is $w_1{w}_2\dots w_k$, It can be processed by logarithm, that is, $\log w_1{w}_2\dots w_k=\log w_1+\log w_2+\dots \log w_k$. In this way, the multiplication is transformed into adding points. In order to find the longest path, the opposite number can be processed. Because $w_i(0\le w_i\le 1)$，所以$\log w_i\le 0$ , all non-negative numbers after taking the opposite number, so Dijkstra algorithm can be used.

Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 2010;

int n,m,S,T;
double g[N][N];
double dist[N];
bool st[N];

void dijkstra()
{
    
    
    dist[S] = 1;
    for(int i=1;i<=n;i++){
    
    
        int t = -1;
        for(int j=1;j<=n;j++){
    
    
            if(!st[j]&&(t==-1||dist[t]<dist[j])){
    
    
                t = j;
            }
        }
        st[t] = true;
        for(int j=1;j<=n;j++){
    
    
            dist[j] = max(dist[j],dist[t]*g[t][j]);
        }
    }
}

int main()
{
    
    
    scanf("%d%d",&n,&m);
    while(m--){
    
    
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        double z = (100.0 - c) / 100;
        g[a][b] = g[b][a] = max(g[a][b], z);
    }
    scanf("%d%d",&S,&T);
    dijkstra();
    printf("%.8f\n",100.0/dist[T]);
    return 0;
}

reward

• To find the longest path, you can reverse the side weight and make the shortest path. In fact, in the process of writing the code, it is not necessary to actually reverse it, just reverse the inequality sign of the triangle inequality, $min$ changed to$mAX$ can.
• For multiplication, if the weight is less than or equal to $1$ , then it is equivalent to the negative side weight of addition; if it is greater than$In$ the case of$1$ , it means the right to the front.