POJ 3662 二分+Dijkstra

题意：


思路：
二分+Disjktra

二分一个值 如果某条边的边权比它小，则连上边权为0的边，否则连上边权为1的边

最后的d[n]就是最小要免费连接多少电话线。

//By SiriusRen
#include 
#include 
#include 
#include 
using namespace std;
#define N 22222
int n,p,k,ans=-1,w[N],first[N],v[N],next[N],tot,d[N],vis[N];
void add(int x,int y,int z){w[tot]=z,v[tot]=y,next[tot]=first[x],first[x]=tot++;}
struct Node{int from,to,weight;}node[11111],jy;
bool operator<(Node a,Node b){return a.weight>b.weight;}
void Dijkstra(int x){
    priority_queuepq;
    memset(d,0x3f,sizeof(d));
    memset(vis,0,sizeof(vis));
    d[1]=0;jy.to=1,jy.weight=0;
    pq.push(jy);
    while(!pq.empty()){
        Node t=pq.top();pq.pop();
        if(!vis[t.to])vis[t.to]=1;
        else continue;
        for(int i=first[t.to];~i;i=next[i]){
            if(!vis[v[i]]&&d[v[i]]>d[t.to]+w[i]){
                d[v[i]]=d[t.to]+w[i];
                jy.to=v[i],jy.weight=d[v[i]];
                pq.push(jy);
            }
        }
    }
}
bool check(int x){
    memset(first,-1,sizeof(first)),tot=0;
    for(int i=1;i<=p;i++)
    {
        if(node[i].weight<=x){
            add(node[i].from,node[i].to,0);
            add(node[i].to,node[i].from,0);
        }
        else
        {
            add(node[i].from,node[i].to,1);
            add(node[i].to,node[i].from,1);
        }
    }
    Dijkstra(x);
    return d[n]<=k;
}
int main(){
    scanf("%d%d%d",&n,&p,&k);
    for(int i=1;i<=p;i++)
        scanf("%d%d%d",&node[i].from,&node[i].to,&node[i].weight);
    int l=0,r=0x3fffffff;
    while(l<=r){
        int mid=(l+r)>>1;
        if(check(mid))r=mid-1,ans=mid;
        else l=mid+1;
    }
    printf("%d\n",ans);
}