题目链接
题意:求一d条1 到 n的路径,使得路径上第k+1大的边权最小
单调性:二分答案k+1条的边权mid 如果mid小了,那越多的边>=mid,
我们最后取到k
#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
int h[N],e[N],w[N],idx,ne[N];
int n,p,k;
typedef pair<int,int>pll;
void add(int a,int b,int c){
e[idx]=b;
w[idx]=c;
ne[idx]=h[a];
h[a]=idx++;
}
int dis[N];
bool vis[N];
int dijistra(int mid){
memset(dis,0x3f,sizeof dis);
memset(vis,0,sizeof vis);
dis[1]=0;
priority_queue<pll,vector<pll>,greater<pll>>q;
q.push({0,1});
while(q.size()){
auto t=q.top();
q.pop();
int ver=t.second,distance=t.first;
if(vis[ver]) continue;
vis[ver]=true;
for(int i=h[ver];i!=-1;i=ne[i]){
int j=e[i],z=w[i];
if(vis[j]) continue;
if(z>=mid) z=1;
else z=0;
if(dis[j]>dis[ver]+z){
dis[j]=dis[ver]+z;
q.push({dis[j],j});
}
}
}
return dis[n];
}
int main(){
cin>>n>>p>>k;
memset(h,-1,sizeof h);
while(p--){
int a,b,l;
cin>>a>>b>>l;
add(a,b,l);
add(b,a,l);
}
int l=0,r=1e6;
if(dijistra(0)>0x3f3f3f3f/2) cout<<"-1"<<endl;
else{
while(r>l){
int mid=l+r+1>>1;
if(dijistra(mid)>k) l=mid;
else r=mid-1;
}
cout<<l<<endl;
}
return 0;
}