题目戳 题目描述 Farmer John wants the cows to prepare for the county jumping competition, so Bessie and the gang are practicing jumping over hurdles. They are getting tired, though, so they want to be able to use as little energy as possible to jump over th…

题目描述 Farmer John wants the cows to prepare for the county jumping competition, so Bessie and the gang are practicing jumping over hurdles. They are getting tired, though, so they want to be able to use as little energy as possible to jump over the hu…

P2888 [USACO07NOV]牛栏Cow Hurdles Floyd $n<=300$?果断Floyd 给出核心式,自行体会 $d[i][j]=min(d[i][j],max(d[i][k],d[k][j]))$ #include<iostream> #include<cstdio> #include<cstring> using namespace std; int max(int a,int b){return a>b?a:b;} int min(…

题目描述 Farmer John wants the cows to prepare for the county jumping competition, so Bessie and the gang are practicing jumping over hurdles. They are getting tired, though, so they want to be able to use as little energy as possible to jump over the hu…

OJ题号:洛谷2888 思路:修改Floyd,把边权和改为边权最大值.另外注意是有向图. #include<cstdio> #include<algorithm> using namespace std; #define inf 0x7fffffff int main() { int n,m,t; scanf("%d%d%d",&n,&m,&t); ][n+]; ;i<=n;i++) { ;j<=n;j++) { d[i][j…

题意很简单,给一张图,把基本的求起点到终点最短路改成求经过k条边的最短路. 求最短路常用的算法是dijkstra,SPFA,还有floyd. 考虑floyd的过程: c[i][j]=min(c[i][j],a[i][k]+b[k][j]); 自然而然联想到矩阵乘法,每次加入一个点就相当于多加一条边,那么加k次就是k条边的最短路. 但是k可能很大(见数据范围),那么显然直接循环矩乘k次是行不通的,于是就想到了矩阵快速幂. 和普通快速幂一样的方式,只不过是把乘法替换成矩乘. 代码如下: #inclu…

题面 解题思路 ## floyd+矩阵快速幂,跟GhostCai爷打赌用不用离散化,最后完败..GhostCai真是tql ! 有个巧妙的方法就是将节点重新编号,因为与节点无关. 代码 #include<bits/stdc++.h> using namespace std; const int MAXN = 1005; int n,t,s,e; int edge[MAXN][MAXN]; int num[MAXN],tot; struct Mat{ int a[105][105]; Mat o…

本题就是求两点间只经过n条边的最短路径,定义广义的矩阵乘法,就是把普通的矩阵乘法从求和改成了取最小值,把内部相乘改成了相加. 代码包含三个内容:广义矩阵乘法,矩阵快速幂,离散化: 1 #include<bits/stdc++.h> 2 using namespace std; 3 const int INF=0x3f3f3f3f; 4 const int N=120; 5 int Hash[1000005],cnt=0;//用于离散化 6 struct matrix{ 7 int m[N][N…

P2887 [USACO07NOV]防晒霜Sunscreen 题目描述 To avoid unsightly burns while tanning, each of the C (1 ≤ C ≤ 2500) cows must cover her hide with sunscreen when they're at the beach. Cow i has a minimum and maximum SPF rating (1 ≤ minSPFi ≤ 1,000; minSPFi ≤ max…

P1472 奶牛家谱 Cow Pedigrees 102通过 193提交 题目提供者该用户不存在 标签USACO 难度普及+/提高 提交  讨论  题解 最新讨论 暂时没有讨论 题目描述 农民约翰准备购买一群新奶牛. 在这个新的奶牛群中, 每一个母亲奶牛都生两个小奶牛.这些奶牛间的关系可以用二叉树来表示.这些二叉树总共有N个节点(3 <= N < 200).这些二叉树有如下性质: 每一个节点的度是0或2.度是这个节点的孩子的数目. 树的高度等于K(1 < K < 100).高度是从…