luogu P1462 通往奥格瑞玛的道路

嘟嘟嘟

这道题的题面相当的迷，我看了半天都没懂。最后看了题解的解释才懂。

他是这么个意思：对于所有能活着走到终点的路径，输出每一条路径中过路费最多的城市的最小值。

那么自然想到二分过路费，然后用dijkstra或spfa判断是否存在一条路径，该路径上的每一个城市的过路费都小于当前二分值mid。用dijkstra复杂度就是O(nlog²n)，spfa玄学，但也能过。

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<vector>
 #include<stack>
 #include<queue>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a, x) memset(a, x, sizeof(a))
 #define rg register
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = 1e4 + ;
 const int maxe = 5e4 + ;
 inline ll read()
 {
   ll ans = ;
   char ch = getchar(), last = ' ';
   while(!isdigit(ch)) last = ch, ch = getchar();
   while(isdigit(ch)) ans = (ans << ) + (ans << ) + ch - '', ch = getchar();
   if(last == '-') ans = -ans;
   return ans;
 }
 inline void write(ll x)
 {
   if(x < ) x = -x, putchar('-');
   if(x >= ) write(x / );
   putchar(x %  + '');
 }

 int n, m;
 ll b, a[maxn], Max = ;

 struct Edge
 {
   int nxt, to; ll w;
 }e[maxe << ];
 int head[maxn], ecnt = -;
 void addEdge(int x, int y, ll w)
 {
   e[++ecnt] = (Edge){head[x], y, w};
   head[x] = ecnt;
 }

 #define pr pair<ll, int>
 #define mp make_pair
 ll dis[maxn];
 bool in[maxn];
 priority_queue<pr, vector<pr>, greater<pr> > q;
 bool dijkstra(ll x)
 {
   for(int i = ; i <= n; ++i) dis[i] = (ll)INF * (ll)INF, in[i] = ;
   dis[] = ;
   while(!q.empty()) q.pop();
   q.push(mp(dis[], ));
   while(!q.empty())
     {
       int now = q.top().second; q.pop();
       if(in[now]) continue;
       in[now] = ;
       for(int i = head[now]; i != -; i = e[i].nxt)
     {
       if(dis[e[i].to] > dis[now] + e[i].w && dis[now] + e[i].w < b && a[e[i].to] <= x)
         {
           if(e[i].to == n) return ;
           dis[e[i].to] = dis[now] + e[i].w;
           q.push(mp(dis[e[i].to], e[i].to));
         }
     }
     }
   return ;
 }

 int main()
 {
   Mem(head, -);
   n = read(); m = read(); b = read();
   for(int i = ; i <= n; ++i) a[i] = read();
   for(int i = ; i <= m; ++i)
     {
       int x = read(), y = read(); ll w = read();
       if(w > b) continue;
       Max = max(Max, w);
       addEdge(x, y, w); addEdge(y, x, w);
     }
   ll L = , R = (ll)INF * (ll)INF;
   while(L < R)
     {
       ll mid = (L + R) >> ;
       if(dijkstra(mid)) R = mid;
       else L = mid + ;
     }
   if(L == (ll)INF * (ll)INF) puts("AFK");
   else write(L), enter;
   return ;
 }