题面 大意:在加权无向图上求出一条从 \(1\) 号结点到 \(N\) 号结点的路径,使路径上第 \(K + 1\) 大的边权尽量小. 思路: 由于我们只能直接求最短路,不能记录过程中的具体的边--那样会特别麻烦 所以,我们就尝试着去想更优的办法 题目中所说,能够免去 \(K\) 条边的费用,那么对于要建设的边中,肯定免去费用最大 $K $ 条边更优 我们关注的应该是第 \(K+1\) 大边的边权,因为其他边权大于这条边边权的边都会被略去 这条边可以枚举吗? 似乎不行,枚举的复杂度还要在最短路的…

传送门 Luogu 解题思路 考虑二分,每次把大于二分值的边的权设为1,小于等于的设为0,如果最短路<=k则可行,记得判无解 细节注意事项 咕咕咕 参考代码 #include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <cctype> #include <cmath>…

P1948 [USACO08JAN]电话线Telephone Lines 题目描述 Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company is uncooperative, so he needs to pay for some of the cables required to connect his farm to the phone system. There a…

P1948 [USACO08JAN]电话线Telephone Lines 题目描述 Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company is uncooperative, so he needs to pay for some of the cables required to connect his farm to the phone system. There a…

P1948 [USACO08JAN]电话线Telephone Lines 题意 题目描述 Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company is uncooperative, so he needs to pay for some of the cables required to connect his farm to the phone system. Ther…

目录 题面 题目链接 题目描述 输入输出格式 输入格式 输出格式 输入输出样例 输入样例 输出样例 说明 思路 AC代码 题面 题目链接 P1948 [USACO08JAN]电话线Telephone Lines 题目描述 Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company is uncooperative, so he needs to pay for some of…

题目描述 Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company is uncooperative, so he needs to pay for some of the cables required to connect his farm to the phone system. There are N (1 ≤ N ≤ 1,000) forlorn telephon…

多年以后,笨笨长大了,成为了电话线布置师.由于地震使得某市的电话线全部损坏,笨笨是负责接到震中市的负责人.该市周围分布着N(1<=N<=1000)根据1……n顺序编号的废弃的电话线杆,任意两根线杆之间没有电话线连接,一共有p(1<=p<=10000)对电话杆可以拉电话线.其他的由于地震使得无法连接. 第i对电线杆的两个端点分别是ai,bi,它们的距离为li(1<=li<=1000000).数据中每对(ai,bi)只出现一次.编号为1的电话杆已经接入了全国的电话网络,整个…

这道题其实是分层图,但和裸的分层图不太一样.因为它只要求路径总权值为路径上最大一条路径的权值,但仔细考虑,这同时也满足一个贪心的性质,那就是当你每次用路径总权值小的方案来更新,那么可以保证新的路径权值尽量小. 所以这道题在不删边的情况下可以使用Dij来跑,而删边权的情况就是分层图. 所以就拿分层图来搞好了^_^. 由于这个数据p和k都比较大,所以直接建k+1层图是要爆的,而k+1层图边都一样,我们就用dis[层数(0-k)]来表示. 具体的就是每次Dij转移是要分两种情况: 1.在原层跑,也就是…

题面 题解 很显然,答案满足单调性. 因此,可以使用二分答案求解. 考虑\(check\)的实现. 贪心地想,免费的\(k\)对电话线一定都要用上. 每次\(check\)时将小于\(mid\)的边权设为\(0\),其它的设为\(1\). 跑一边最短路判断\(\mathrm{dist[n]}\)是否\(\leq k\)即可. 代码 #include <bits/stdc++.h> #define itn int #define gI gi using namespace std; inline…