题目 两眼题 二分一个\(lim\),然后跑最短路(边权\(\le lim\)的边长度为\(0\),\(>lim\)的长度为\(1\)),然后判断\(dis_{1,n}\le k\). #include<bits/stdc++.h> #define pi pair<int,int> #define pb push_back using namespace std; int read(){int x;scanf("%d",&x);return x;}…

题意描述 在无向图中求一条从 \(1\) 到 \(N\) 的路径,使得路径上第 \(K+1\) 大的边权最小. 等等,最大的最小...如此熟悉的字眼,难道是 二分答案. 下面进入正题. 算法分析 没错就是酱紫,二分这个第 \(K+1\) 大的边权的值,设这条边的边权为 \(val\) 那么: \[ edge[i]=\left\{ \begin{array}{rcl} 0 & & {edge[i] \leq val}\\ 1 & & {edge[i] > val}\ \…

嘟嘟嘟 题意概括一下,就是在无向图上求一条1到n的路径,使路径上第k + 1大的边权尽量小. 考虑dp,令dp[i][j] 表示走到节点 i,路线上有 j 条电话线免费时,路径上最贵的电缆花费最小是多少.则对于一条从u到v,长度为w的边,转移方程是: 1.这条电缆要付费:dp[v][p] = min(dp[v][p], max(dp[u][p], w)) 2.这条电缆免费:dp[v][p + 1] = min(dp[v][p +1], dp[u][p]) 不过这是在图上dp,转移不能保证无后效性…

这道题目有两种解法: 1.将每个点视为一个二元组(x,p),表示从起点到x有p条路径免费,相当于构建了一张分层图,N*k个节点,P*k条边.在这张图上用优先队列优化的SPFA算法求解,注意这里的d数组存的不是最短路径,而是路径中边权最大的值,最终答案就是min(d[n][j]),0<=j<=k . 1 #include<bits/stdc++.h> 2 using namespace std; 3 const int N=1005,M=20005; 4 int head[N],to…

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…

洛谷 P1948 [USACO08JAN]电话线Telephone Lines https://www.luogu.org/problem/P1948 JDOJ 2556: USACO 2008 Jan Silver 3.Telephone Lines https://neooj.com/oldoj/problem.php?id=2556 题目描述 Farmer John wants to set up a telephone line at his farm. Unfortunately, t…

1614: [Usaco2007 Jan]Telephone Lines架设电话线 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 892  Solved: 399[Submit][Status] Description Farmer John打算将电话线引到自己的农场,但电信公司并不打算为他提供免费服务.于是,FJ必须为此向电信公司支付一定的费用. FJ的农场周围分布着N(1 <= N <= 1,000)根按1..N顺次编号的废弃的电话线杆,任意两…

POJ3662 Telephone Lines 题目大意:要在顶点1到顶点n之间建一条路径,假设这条路径有m条边,其中有k条边是免费的,剩余m-k条边是要收费的, 求这m-k条边中花费最大的一条边的最小花费. 让m条边中原本花费最大的k条边成为免费的边,则这时m-k条边中花费最大的一条边的花费最小. 二分枚举m-k条边中花费最大的一条边的最小花费x,dijkstra求最短路径时,将花费大于x的边的花费设为1(花费为INF的边不变),花费小于等于x的边设为 0,则d[v-1]中返回的就是花费大于x…

题目 1614: [Usaco2007 Jan]Telephone Lines架设电话线 Time Limit: 5 Sec  Memory Limit: 64 MB Description Farmer John打算将电话线引到自己的农场,但电信公司并不打算为他提供免费服务.于是,FJ必须为此向电信公司支付一定的费用. FJ的农场周围分布着N(1 <= N <= 1,000)根按1..N顺次编号的废弃的电话线杆,任意两根电话线杆间都没有电话线相连.一共P(1 <= P <= 10…

Telephone Lines Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7115   Accepted: 2603 Description 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…

Telephone Lines Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7214   Accepted: 2638 Description 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…

题目链接:http://poj.org/problem?id=3662 Telephone Lines Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8248   Accepted: 2977 Description Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company is uncoopera…

http://poj.org/problem?id=3662 Telephone Lines Time Limit: 1000MS   Memory Limit: 65536K Total Submissions:9310   Accepted: 3374 Description Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company is uncooperative,…

Telephone Lines Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6785   Accepted: 2498 Description 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…

Telephone Lines Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7978 Accepted: 2885 Description 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…

Telephone Lines Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6973   Accepted: 2554 Description 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…

Telephone Lines bzoj-1614 Usaco-2007Jan 题目大意:给你一个n个点m条边的带边权无向图,求最短路.可以选取k条边免费. 注释:$1\le n\le 10^3$,$1\le m\le 10^5$ 想法:一眼分层图最短路啊! 我都想出来了就上网查一下题解吧 卧槽??二分+spfa?? 其实这个题不用分层图 我们二分答案,二分出最大值,然后将图中所有比mid大的边设为1,小的设为0,然后跑最短路.如果最短路的值比k大,说明最大值一定比二分的mid大.因为我的最短路…

Telephone Lines Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5696   Accepted: 2071 Description 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…

[Usaco2007 Jan]Telephone Lines架设电话线 Description FarmerJohn打算将电话线引到自己的农场,但电信公司并不打算为他提供免费服务.于是,FJ必须为此向电信公司支付一定的费用.FJ的农场周围分布着N(1<=N<=1,000)根按1..N顺次编号的废弃的电话线杆,任意两根电话线杆间都没有电话线相连.一共P(1<=P<=10,000)对电话线杆间可以拉电话线,其余的那些由于隔得太远而无法被连接. 第i对电话线杆的两个端点分别为A_i.B_…

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…

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…

思路 考虑题目要求求出最小的第k+1大的边权,想到二分答案 然后二分第k+1大的边权wx 把所有边权<=wx的边权变为0,边权>wx的边权变为0,找出最短路之后,如果dis[T]<=k,则答案可行,反之则不可行 似乎有dp解法的样子,真神奇 代码 #include <cstdio> #include <algorithm> #include <cstring> #include <queue> using namespace std; in…

传送门 思路: 二分+最短路径:可以将长度小于等于 mid 的边视为长度为 0 的边,大于 mid 的边视为长度为 1 的边,最后用 dijkstra 检查 d [ n ] 是否小于等于 k 即可. 标程: #include<cstring> #include<queue> #include<cstdio> #include<iostream> #include<vector> #include<fstream> #include&l…

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…

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…

#\(\mathcal{\color{red}{Description}}\) \(Link\) 给定一个图,请你求出在把其中自由选择的\(k\)条的权值都置为零的情况下,图中\(1-N\)最短路上的最大边权的最小值. #\(\mathcal{\color{red}{Solution}}\) 哇这个题真是吊打我的智商啊- 首先我们看题目中给的限制条件,限制我们不能直接\(sort\)一遍的条件就是我们要找的是最短路上的边权最大值最小限制了我们把一些边的权值置为零之后,图上的最短路.而这个最短路的…

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…