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Cow relays

给定一张无向图 求S->T经过K条边的最短路是多少 对于形如 \(A(i,j)=opt(A(i,k)\ opt\ B(k,j)\)的形式 可以考虑用矩阵乘法来解决 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define ll long long using namespace std; const int N=110;

Poj 3613 Cow Relays (图论)

Poj 3613 Cow Relays (图论) 题目大意 给出一个无向图,T条边,给出N,S,E,求S到E经过N条边的最短路径长度 理论上讲就是给了有n条边限制的最短路 solution 最一开始想到是的去直接统计最短路经过了多少条边,结果,,, 还是太年轻了。。。 不过,看数据范围只有1000,那么floyd是首选 回顾Floyd算法流

[USACO07NOV]牛继电器Cow Relays

https://www.luogu.org/problem/P2886 题目描述: 给出一张无向连通图,求$S$到$E$经过$k$条边的最短路。 对于一类$S$到$E$走指定数量的边数,求它的最短路或条数,都可以采用矩阵快速幂的方式解决.我们回忆一下那一个慢得惊人的$floyd$算法,将它的$dp$方程式提取出来. $floyd[i][j]=m

relays 奶牛接力跑(矩阵快速幂求最短路径)

题干:   FJ的N(2<=N<=1,000,000)头奶牛选择了接力跑作为她们的日常锻炼项目。至于进行接力跑的地点 自然是在牧场中现有的T(2 <= T <= 100)条跑道上。农场上的跑道有一些交汇点,每条跑道都连结了两个不同的交汇点 I1_i和I2_i(1<=I1_i<=1,000; 1<=I2_i<=1,000)。每个交汇点都是至少

poj3613 Cow Relays

题目 题目传送门 Description For their physical fitness program, N (2 ≤ N ≤ 1,000,000) cows have decided to run a relay race using the T (2 ≤ T ≤ 100) cow trails throughout the pasture. Each trail connects two different intersections (1 ≤ I1i ≤

POJ3613--Cow Relays

Description For their physical fitness program, N (2 ≤ N ≤ 1,000,000) cows have decided to run a relay race using the T (2 ≤ T ≤ 100) cow trails throughout the pasture. Each trail connects two different intersections (1 ≤ I1i ≤ 1,000; 1 ≤ I2

[POJ-3613]Cow Relays

题目传送门 解析:这道题本质上是求两点之间经过n条边的最短路,是一道快速幂的应用,只需把快速幂自乘的操作改成最短路边数*2,把快速幂乘答案的操作改成维护答案即可。 #include<iostream>#include<cstdio>#include<map>#include<cstring>using namespace std;int minn[1005][1005],b[1

POJ1966 Cable TV Network

Cable TV Network Language:DefaultCable TV Network Time Limit: 1000MSMemory Limit: 30000KTotal Submissions: 5273Accepted: 2446DescriptionThe interconnection of the relays in a cable TV network is bi-directional. The network is connected if there is at lea

Cable TV Neteork

Description The interconnection of the relays in a cable TV network is bi-directional. The network is connected if there is at least one interconnection path between each pair of relays present in the network. Otherwise the network is disconnected. An emp