思路:要最短时间全部吧信送完就是找最短路中最长花费的时间是多久
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define SIS std::ios::sync_with_stdio(false)
#define space putchar(' ')
#define enter putchar('\n')
#define lson root<<1
#define rson root<<1|1
typedef pair<int,int> PII;
const int mod=1e9+7;
const int N=1e5+5;
const int inf=0x7f7f7f7f;
int gcd(int a,int b)
{
return b==0?a:gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
return a*(b/gcd(a,b));
}
template <class T>
void read(T &x)
{
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-')
op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op)
x = -x;
}
template <class T>
void write(T x)
{
if(x < 0)
x = -x, putchar('-');
if(x >= 10)
write(x / 10);
putchar('0' + x % 10);
}
ll qsm(int a,int b,int p)
{
ll res=1%p;
while(b)
{
if(b&1)
res=res*a%p;
a=1ll*a*a%p;
b>>=1;
}
return res;
}
struct node
{
int to,nex,w;
}edge[N];
int tot=0,head[N];
int dis[N];
int vis[N];
int n,m;
void add(int u,int v,int w)
{
tot++;
edge[tot].nex=head[u];
edge[tot].to=v;
edge[tot].w=w;
head[u]=tot;
}
int dj(int ts)
{
memset(dis,inf,sizeof dis);
memset(vis,0,sizeof vis);
priority_queue<PII,vector<PII>,greater<PII> > q;
q.push({
0,ts});
dis[ts]=0;
while(!q.empty())
{
auto now=q.top();
q.pop();
int d=now.first,u=now.second;
if(vis[u])continue;
vis[u]=1;
for(int i=head[u];~i;i=edge[i].nex)
{
int v=edge[i].to;
if(dis[v]>d+edge[i].w)
{
dis[v]=d+edge[i].w;
q.push({
dis[v],v});
}
}
}
int ans=-inf;
for(int i=1;i<=n;i++){
if(dis[i]==inf)return -1;
ans=max(ans,dis[i]);
}
return ans;
}
int main()
{
memset(head,-1,sizeof head);
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
add(u,v,w);
add(v,u,w);
}
printf("%d", dj(1));
return 0;
}