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…
Sunscreen Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9333 Accepted: 3264 Description 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…
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…
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…
