dijkstra题解

将dis[i]设为到第i顶点最多可以保留百分之多少，最后相除再*100即可，详见代码

#include<bits/stdc++.h>
using namespace std;

int n,m,A,B;

struct Node{
	int v;
	double val;
	friend bool operator < (Node a,Node b){
		return a.val<b.val;//保留的越多越好，所以要从大到小排序
	}
};

struct G_data{
	int v;
	double val;
};

priority_queue<Node> q;
vector<G_data> G[2005];
double dis[2005];
bool vis[2005];
const int INF=1e9+7;

void dij(){
	for(int i=0;i<=n+1;i++){
		dis[i]=0;
	}
	dis[A]=100;
	q.push({A,0});
	while(!q.empty()){
		Node f=q.top();
		q.pop();
		vis[f.v]=true;
		
		for(int i=0;i<G[f.v].size();i++){
			int pos=G[f.v][i].v;
			double w=G[f.v][i].val;
			//cout<<"-"<<pos<<endl;
			if(dis[pos]<w*dis[f.v]){
				dis[pos]=w*dis[f.v];
				if(vis[pos])continue;
				q.push({pos,dis[pos]});
			}
		}
	}
}


int main(){
	cin>>n>>m;
	while(m--){
		int u,v;
		double w;
		cin>>u>>v>>w;
		double nw_W=(100-w)/100.0;//可留下百分之几
		G[u].push_back({v,nw_W});
		G[v].push_back({u,nw_W});
	}
	cin>>A>>B;
	dij();
	//for(int i=1;i<=n;i++)cout<<dis[i]<<endl;
	double ans=100.0/dis[B]*100.0;//将100补乘回来
	printf("%.8f\n", ans);
}