【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=0,f=1;
    char ch=getchar();
    while(!isdigit(ch)){
        if(ch=='-')    f=-1;
        ch=getchar();
    }
    while(isdigit(ch)){
        num=num*10+ch-'0';
        ch=getchar();
    }
    return num*f;
}

struct Edge{
    int next,to,dis;
}edge[1000010];
int head[100100],num;
inline void add(int from,int to,int dis){
    edge[++num]=(Edge){head[from],to,dis};
    head[from]=num;
}

int val[100010];
int f[1000001],h,t=1;
bool vis[100010];
int dis[100010];
int ans;

int main(){
    int n=read(),m=read(),k=read();
    for(int i=1;i<=m;++i){
        int from=read(),to=read(),dst=read();
        add(from,to,dst);
        add(to,from,dst);
        val[i]=dst;
    }
    std::sort(val+1,val+m+1);
    int l=1,r=m;
    
    while(l<=r){
        int mid=(l+r)>>1;
        int cost=val[mid];
        memset(vis,0,sizeof(vis));
        memset(dis,127/3,sizeof(dis));
        f[1]=1;h=0;t=1;dis[1]=0;
        while(h++<t){
            int from=f[h];
            vis[from]=0;
            for(int i=head[from];i;i=edge[i].next){
                int to=edge[i].to;
                int c;
                if(edge[i].dis<=cost)    c=0;
                else                     c=1;
                if(dis[to]>dis[from]+c){
                    dis[to]=dis[from]+c;
                    if(!vis[to]){
                        vis[to]=1;
                        f[++t]=to;
                    }
                }
            }
        }
        if(dis[n]>k)    l=mid+1;
        else{
            ans=cost;
            r=mid-1;
        }
    }
    printf("%d",ans==0?-1:ans);
    return 0;
}