[bzoj1614][Usaco2007Jan]Telephone Lines 架设电话线_二分答案_最短路

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#define N 1000+50
#define M 20000+50
using namespace std;
int head[N],dis[N];
const int inf = 0x7fffff;
int pre[N],from[N];
struct graph
{
    int next,to,val;
    int deval;
    graph () {}
    graph (int _next,int _to,int _val)
    :next(_next),to(_to),val(_val){}
}edge[M];
int cnt = 0;
inline void add(int x,int y,int z)
{
    edge[++cnt] = graph(head[x],y,z);
    head[x]=cnt;
    edge[++cnt] = graph(head[y],x,z);
    head[y]=cnt;
}
int S,T;
bool in[N];
queue<int>Q;
void spfa()
{
    memset(dis,0x3f,sizeof dis);
    memset(in,false,sizeof in);
    memset(pre,0,sizeof pre);
    memset(from,0,sizeof from);
    Q.push(S),dis[S]=0;in[S]=1;
    while(!Q.empty())
    {
        int tt=Q.front();Q.pop();in[tt]=false;
        for(int i=head[tt];i;i=edge[i].next)
            if(dis[edge[i].to]>dis[tt]+edge[i].deval)
            {
                dis[edge[i].to]=dis[tt]+edge[i].deval;
                pre[edge[i].to] = tt;from[edge[i].to] = i;
                if(!in[edge[i].to])
                {
                    Q.push(edge[i].to);
                    in[edge[i].to]=1;
                }
            }
    }
}
int n,m,k;
bool check(int mid)
{
    for(int i=1;i<=cnt;++i)
    {
        if(edge[i].val<=mid)edge[i].deval = 0;
        else edge[i].deval = 1;
    }
    spfa();
    if(dis[T]>k)
        return false;
    return true;
}
inline int read()
{
    int x=0,f=1;char ch = getchar ();
    while (ch < ‘0‘ || ch > ‘9‘) {if (ch == ‘-‘)f=-1;ch = getchar();}
    while (ch >=‘0‘ && ch <=‘9‘) {x=(x<<1)+(x<<3)+ch -‘0‘;ch = getchar();}
    return x*f;
}
int main()
{   
    n=read(),m=read(),k=read();
    S = 1,T = n;
    int l = 0 , r = 0;
    for(int i=1;i<=m;++i)
    {
        int x=read(),y=read(),z=read();
        add(x,y,z); l = min(l,z),r = max(r,z);
    }
    int ans = inf;
    while(l<r)
    {
        int mid = (l+r)>>1;
        if(check(mid))
        {
            int now = n;    
            int tmp = 0;
            while(now ^ 1)
            {
                if(edge[from[now]].deval == 0)
                    tmp = max(tmp,edge[from[now]].val);
                now = pre[now];
            }
            ans = min(ans,tmp);
            r = mid;
        }
        else
            l = mid + 1;
    }
    printf("%d\n",ans == inf ? -1 : ans);
}