POJ3662 Telephone Lines

二分一个长度，(count)dis[n]<=k

O(PlogNlogK)

#include <queue>
#include <iostream> 
using namespace std;
const int INF=1e9,N=1e5+10;
struct edge {
    int v,w;
    edge(int v,int w):v(v),w(w){}
    bool operator<(const edge &a)const{return w>a.w;}
};

vector<edge>g[N];
int n,m,k,dis[N],vis[N];

bool ok(int limit) {
    for(int i=1; i<=n; i++)dis[i]=INF,vis[i]=0;
    priority_queue<edge>q;
    dis[1]=0,q.push(edge(1,0));
    while(!q.empty()) {
        edge e=q.top();q.pop();
        int u=e.v;
        if(vis[u])continue;
        vis[u]=1;
        for(int i=0;i<g[u].size();i++) {
            int&v=g[u][i].v;
            int w=g[u][i].w>limit?1:0;
            if(dis[v]>dis[u]+w) {
                dis[v]=dis[u]+w;
                q.push(edge(v,dis[v]));
            }
        }
    }
    return dis[n]<=k;
}

int solve() {
    int l=0,r=1e7,ans=-1;
    while(l<=r) {
        int mid=(l+r)>>1;
        if(ok(mid))r=mid-1,ans=mid;
        else l=mid+1;
    }
    return ans;
}

int main() {
    while(cin>>n>>m>>k) {
        for(int i=1; i<=n; i++)g[i].clear();
        for(int i=1; i<=m; i++) {
            int u,v,w;
            cin>>u>>v>>w;
            g[u].push_back(edge(v,w));
            g[v].push_back(edge(u,w));
        }
        cout<<solve()<<endl;
    }
    return 0;
}