POJ-3662 Telephone Lines---二分+最短路+最小化第k+1大

题目链接：

https://cn.vjudge.net/problem/POJ-3662

题目大意：

求一条路径从1到n使第k+1大的边最小。

解题思路：

二分答案mid，当原边权小于等于mid新边权为0，否则新边权为1.

求最短路，若小于等于k说明满足条件

 

注意：最开始的l必须是0，而不是这些边中的最小边，因为最小化第k+1大的话答案可能为0

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 #include<queue>
 #define MID(l, r) (l + (r - l) / 2)
 #define lson(o) (o * 2)
 #define rson(o) (o * 2 + 1)
 using namespace std;
 typedef long long ll;
 const int INF = 1e9 +;
 const int maxn = 1e6 + ;
 int n, m, k;
 struct edge{
     int u, v, w;
     edge(int u, int v, int w):u(u), v(v), w(w){}
     edge(){}
 };
 struct Heapnode
 {
     int d, u;
     Heapnode(){}
     Heapnode(int d, int u):d(d), u(u){}
     bool operator <(const Heapnode & a)const
     {
         return d > a.d;
     }
 };
 bool v[maxn];
 int d[maxn];

 int dijkstra(int s, int t, vector<edge>edges, vector<int>G[])
 {
     priority_queue<Heapnode>q;
     for(int i = ; i <= n; i++)d[i] = INF;
     d[s] = ;
     memset(v, , sizeof(v));
     q.push(Heapnode(, s));
     while(!q.empty())
     {
         Heapnode now = q.top();
         q.pop();
         int u = now.u;
         if(v[u])continue;
         v[u] = ;
         for(int i = ; i < G[u].size(); i++)
         {
             edge& e = edges[G[u][i]];
             int v = e.v;
             if(d[v] > d[u] + e.w)
             {
                 d[v] = d[u] + e.w;
                 q.push(Heapnode(d[v], v));
             }
         }
     }
     return d[t];
 }

 vector<edge>edges;
 vector<int>G[maxn];
 vector<edge>edges2;
 void addedge(int u, int v, int w)
 {
     edges.push_back(edge(u, v, w));
     int m = edges.size();
     G[u].push_back(m - );
 }
 void init(int mid)
 {
     edges2.clear();
     for(int i = ; i < edges.size(); i++)
     {
         edge& e = edges[i];
         if(e.w <= mid)
         {
             edges2.push_back(edge(e.u, e.v, ));
         }
         else
         {
             edges2.push_back(edge(e.u, e.v, ));
         }
     }
 }
 int main()
 {
     scanf("%d%d%d", &n, &m, &k);
     int l = , r = , u, v, w;
     while(m--)
     {
         scanf("%d%d%d", &u, &v, &w);
         addedge(u, v, w);
         addedge(v, u, w);
         r = max(r, w);
     }
     int ans = -;
     while(l <= r)//l最开始要设置成0，不能设置成最小边，因为第k大可能为0
     {
         int mid = (l + r) / ;
         init(mid);
         int t = dijkstra(, n, edges2, G);
         //cout<<mid<<" "<<t<<endl;
         if(t <= k)
         {
             ans = mid;
             r = mid - ;
         }
         else
         {
             l = mid + ;
         }
     }
     cout<<ans<<endl;
     return ;
 }