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…

题意:一张带权无向图中,有K条边可以免费修建.现在要修建一条从点1到点N的路,费用是除掉免费的K条边外,权值最大的那条边的值,求最小花费. 分析:假设存在一个临界值X,小于X的边全部免费,那么此时由大于等于X的边组成的图,从点1到点N走过的边数小于等于K,那么这个X就是所求的答案.所以可以通过二分答案的方法求解该问题,每一次根据mid值跑迪杰斯特拉,d[i]记录路径长度(走过边的数目).需要注意的是,要特判一下点1到点N本身不连通的情况以及花费为0的情况.二分的时候,当d[N]>K时修改答案为m…

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: 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…

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 telephone pol…

链接:洛谷 POJ 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) forlor…

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…

题目大意:给定一个 N 个顶点,M 条边的无向图,求一条从 1 号节点到 N 号节点之间的路径,使得第 K+1 大的边权最小,若 1 与 N 不连通,输出 -1. 最小化最大值一类的问题,采用二分答案即可,每次跑一遍 dij ,若边权大于二分的值,那么等效边权为1,否则边权为0,最后判断从 1 到 N 之间的最短路是否大于 K 即可. 代码如下 #include <cstdio> #include <algorithm> #include <memory.h> #inc…

题目传送门 题意:有n个点, p条路,每条道路有个花费Li, 然后现在要建一条1-n的路线,然后可以选k条道路免费, 然后可以在剩下的道路中选择价格最高的边支付费用, 求这个答案最小. 题解: 二分答案. 每次check过程中, 一条边的花费 <= mid 则 路径长度为0,否者路径长度为1.然后 求 到n的点之后长度<=k. 然后就是bfs的过程中. 如果这条边是0, 那么从前入队, 否者从后入队. 代码: #include<cstdio> #include<algorit…

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 telephone pol…