Luogu P1073 Optimal Trade (shortest path)

P1073 Optimal trade

There are many troublesome methods on the Internet, here is the simplest one

设$di{s}_i$Represents $1$ arrivaliiThe smallest point weight (crystal ball price) on all paths of$i$ ,$dis{t}_i$Display $i$ tonnThe maximum point weight on all paths of$n$ .
So, the final answer ismax ⁡ i = 1 n {disti − disi} \max\limits_{i=1}^n \{dist_i-dis_i\}i=1maxn​{ disti​−disi​}

Because each point can go through multiple times, SPFA is used

#include<cstdio>
#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
const int Maxn=100000+10,inf=0x3f3f3f3f;
vector <int> e[Maxn],g[Maxn];
int a[Maxn],dis[Maxn],dist[Maxn];
int n,m,ans;
bool vis[Maxn];
inline int read()
{
    
    
	int s=0,w=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
    
    if(ch=='-')w=-1;ch=getchar();}
	while(ch>='0' && ch<='9')s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
	return s*w;
}
void spfa1()
{
    
    
	fill(dis+1,dis+1+n,inf);
	queue <int> q;
	vis[1]=1,dis[1]=a[1];
	q.push(1);
	while(q.size())
	{
    
    
		int x=q.front();
		q.pop();
		vis[x]=0;
		for(int i=0;i<e[x].size();++i)
		{
    
    
			int y=e[x][i];
			if(dis[y]>min(dis[x],a[y]))
			{
    
    
				dis[y]=min(dis[x],a[y]);
				if(!vis[y])vis[y]=1,q.push(y);
			}
		}
	}
}
void spfa2()
{
    
    
	fill(dist+1,dist+1+n,-inf);
	queue <int> q;
	vis[n]=1,dist[n]=a[n];
	q.push(n);
	while(q.size())
	{
    
    
		int x=q.front();
		q.pop();
		vis[x]=0;
		for(int i=0;i<g[x].size();++i)
		{
    
    
			int y=g[x][i];
			if(dist[y]<max(dist[x],a[y]))
			{
    
    
				dist[y]=max(dist[x],a[y]);
				if(!vis[y])vis[y]=1,q.push(y);
			}
		}
	}
}
int main()
{
    
    
	n=read(),m=read();
	for(int i=1;i<=n;++i)
	a[i]=read();
	for(int i=1;i<=m;++i)
	{
    
    
		int x=read(),y=read(),opt=read();
		e[x].push_back(y);
		g[y].push_back(x);
		if(opt==2)
		{
    
    
			e[y].push_back(x);
			g[x].push_back(y);
		}
	}
	spfa1();
	spfa2();
	
	for(int i=1;i<=n;++i)
	ans=max(ans,dist[i]-dis[i]);
	printf("%d\n",ans);
	
	return 0;
}