【Luogu】P1948电话线（二分SPFA）

　　题目链接

　　二分最长的电话线长度。把所有大于这个长度的边权设成1，小于等于的设成零，然后跑SPFA看dis[n]是否>k。若>k则l=mid+1

　　否则r=mid-1

　　放代码

#include<cstdio>
#include<cctype>
#include<cstring>
#include<algorithm>
inline long long read(){
    long long num=,f=;
    char ch=getchar();
    while(!isdigit(ch)){
        if(ch=='-')    f=-;
        ch=getchar();
    }
    while(isdigit(ch)){
        num=num*+ch-'';
        ch=getchar();
    }
    return num*f;
}

struct Edge{
    int next,to,dis;
}edge[];
int head[],num;
inline void add(int from,int to,int dis){
    edge[++num]=(Edge){head[from],to,dis};
    head[from]=num;
}

int val[];
int f[],h,t=;
bool vis[];
int dis[];
int ans;

int main(){
    int n=read(),m=read(),k=read();
    for(int i=;i<=m;++i){
        int from=read(),to=read(),dst=read();
        add(from,to,dst);
        add(to,from,dst);
        val[i]=dst;
    }
    std::sort(val+,val+m+);
    int l=,r=m;

    while(l<=r){
        int mid=(l+r)>>;
        int cost=val[mid];
        memset(vis,,sizeof(vis));
        memset(dis,/,sizeof(dis));
        f[]=;h=;t=;dis[]=;
        while(h++<t){
            int from=f[h];
            vis[from]=;
            for(int i=head[from];i;i=edge[i].next){
                int to=edge[i].to;
                int c;
                if(edge[i].dis<=cost)    c=;
                else                     c=;
                if(dis[to]>dis[from]+c){
                    dis[to]=dis[from]+c;
                    if(!vis[to]){
                        vis[to]=;
                        f[++t]=to;
                    }
                }
            }
        }
        if(dis[n]>k)    l=mid+;
        else{
            ans=cost;
            r=mid-;
        }
    }
    printf("%d",ans==?-:ans);
    return ;
}